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== 10.10#h(0.6em)Working with Radicals


    Sometimes radical notation is more convenient to use than exponents. In these cases, we usually simplify radical expressions algebraically as much as possible before using a calculator to obtain decimal approximations.


    
      
=== Properties of Radicals


      Because #math.equation(block: false, alt: "the nth root of a equals a to the power 1 / n")[$root(n, a) = a^(1 / n)$], we can use the laws of exponents to derive two important properties that are useful in simplifying radicals.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Radicals]


        + #math.equation(block: false, alt: "the nth root of a b equals the nth root of a the nth root of b , for a , b greater than or equal to 0")[$display(root(n, a b) = root(n, a) root(n, b) ", " #hide($b l a n k 000$) "for " a "," b ≥ 0)$]
+ #math.equation(block: false, alt: "the nth root of the fraction a over b equals the fraction the nth root of a over the nth root of b , for a greater than or equal to 0 , b greater than 0")[$display(root(n, frac(a, b)) = frac(root(n, a), root(n, b)) ", " #hide($b l a n k b l a n k$) "for " a ≥ 0 "," " " " " b > 0)$]
]


      As examples, you can verify that


      #math.equation(block: true, alt: "the square root of 36 equals the square root of 4 the square root of 9 and the cube root of the fraction 1 over 8 equals the fraction the cube root of 1 over the cube root of 8")[$sqrt(36) = sqrt(4) sqrt(9) " " " " " " " and " " " " " " " root(3, frac(1, 8)) = frac(root(3, 1), root(3, 8))$]
      #examplebox("Example 1")[][
Which of the following are true?


          + Is #math.equation(block: false, alt: "the square root of 36 plus 64 equals the square root of 36 plus the square root of 64 ?")[$sqrt(36 + 64) = sqrt(36) + sqrt(64) "?"$]
+ Is #math.equation(block: false, alt: "the cube root of 8 open parenthesis 64 close parenthesis equals the cube root of 8 the cube root of 64 ?")[$root(3, 8 ( 64 )) = root(3, 8) root(3, 64) "?"$]
+ Is #math.equation(block: false, alt: "the square root of x squared plus 4 equals x plus 2 ?")[$sqrt(x^(2) + 4) = x + 2 "?"$]
+ Is #math.equation(block: false, alt: "the cube root of 8 x cubed equals 2 x ?")[$root(3, 8 x^(3)) = 2 x "?"$]


        
        #solutionbox[
The statements in (b) and (d) are true, and both are examples of the first property of radicals.


          Statements (a) and (c) are false. In general, #math.equation(block: false, alt: "the nth root of a plus b")[$root(n, a + b)$] is not equal to #math.equation(block: false, alt: "the nth root of a plus the nth root of b")[$root(n, a) + root(n, b)$], and #math.equation(block: false, alt: "the nth root of a minus b")[$root(n, a − b)$] is not equal to #math.equation(block: false, alt: "the nth root of a minus the nth root of b")[$root(n, a) − root(n, b)$].
]
]


    
    
      
=== Simplifying Radicals


      We use Property (1) to simplify radical expressions by factoring the radicand. For example, to simplify #math.equation(block: false, alt: "the cube root of 108")[$root(3, 108)$], we look for perfect cubes that divide evenly into #math.equation(block: false, alt: "108")[$108$]. The easiest way to do this is to try the perfect cubes in order:


      #math.equation(block: true, alt: "1 , 8 , 27 , 64 , 125 , and so on")[$1 , " " 8 , " " 27 , " " 64 , " " 125 , …$]
      and so on, until we find one that is a factor. For this example, we find that #math.equation(block: false, alt: "108 equals 27 times 4")[$108 = 27 ⋅ 4$]. Using Property (1), we write


      #math.equation(block: true, alt: "the cube root of 108 equals the cube root of 27 the cube root of 4")[$root(3, 108) = root(3, 27) root(3, 4)$]
      Simplify the first factor to find


      #math.equation(block: true, alt: "the cube root of 108 equals 3 the cube root of 4")[$root(3, 108) = 3 root(3, 4)$]
      This expression is considered simpler than the original radical because the new radicand, #math.equation(block: false, alt: "4")[$4$], is smaller than the original, #math.equation(block: false, alt: "108")[$108$].


      We can also simplify radicals containing variables. If the exponent on the variable is a multiple of the index, we can extract the variable from the radical. For instance,


      #math.equation(block: true, alt: "the cube root of 12 equals x to the power 12 / 3 equals x to the power 4")[$root(3, 12) = x^(12 / 3) = x^(4)$]
      (You can verify this by noting that #math.equation(block: false, alt: "open parenthesis x to the power 4 close parenthesis cubed equals x to the power 12")[$( x^(4) )^(3) = x^(12)$].) If the exponent on the variable is not a multiple of the index, we factor out the highest power that is a multiple. For example,


      #math.equation(block: true, alt: "the cube root of x to the power 11, equals the cube root of x to the power 9 times x squared, Apply Property (1).; equals the cube root of x to the power 9 times the cube root of x squared, Simplify the cube root of x to the power 9 equals x to the power 9 / 3.; equals x cubed the cube root of x squared")[$root(3, x^(11)) & = root(3, x^(9) ⋅ x^(2)) &  & "Apply Property (1)." \  & = root(3, x^(9)) ⋅ root(3, x^(2)) &  & "Simplify " root(3, bold(italic(x))^(9)) = bold(italic(x))^(9 / 3) . \  & = x^(3) root(3, x^(2))$]
      #examplebox("Example 2")[][
Simplify each radical.


          + #math.equation(block: false, alt: "the square root of 18 x to the power 5")[$sqrt(18 x^(5))$]
+ #math.equation(block: false, alt: "the cube root of 24 x to the power 6 y to the power 8")[$root(3, 24 x^(6) y^(8))$]


        
        #solutionbox[
+ The index of the radical is #math.equation(block: false, alt: "2")[$2$], so we look for perfect square factors of #math.equation(block: false, alt: "18 x to the power 5")[$18 x^(5)$]. The factor #math.equation(block: false, alt: "9")[$9$] is a perfect square, and #math.equation(block: false, alt: "x to the power 4")[$x^(4)$] has an exponent divisible by #math.equation(block: false, alt: "2")[$2$]. Thus,
                #math.equation(block: true, alt: "the square root of 18 x to the power 5, equals the square root of 9 x to the power 4 times 2 x, Apply Property (1).; equals the square root of 9 x to the power 4 the square root of 2 x, Take square roots.; equals 3 x squared the square root of 2 x")[$sqrt(18 x^(5)) & = sqrt(9 x^(4) ⋅ 2 x) &  & "Apply Property (1)." \  & = sqrt(9 x^(4)) sqrt(2 x) &  & "Take square roots." \  & = 3 x^(2) sqrt(2 x)$]
+ The index of the radical is #math.equation(block: false, alt: "3")[$3$], so we look for perfect cube factors of #math.equation(block: false, alt: "24 x to the power 6 y to the power 8")[$24 x^(6) y^(8)$]. The factor #math.equation(block: false, alt: "8")[$8$] is a perfect cube, and #math.equation(block: false, alt: "x to the power 6")[$x^(6)$] and #math.equation(block: false, alt: "y to the power 6")[$y^(6)$] have exponents divisible by #math.equation(block: false, alt: "3")[$3$]. Thus,
                #math.equation(block: true, alt: "the cube root of 24 x to the power 6 y to the power 8, equals the cube root of 8 x to the power 6 y to the power 6 times 3 y squared, Apply Property (1).; equals the cube root of 8 x to the power 6 y to the power 6 the cube root of 3 y squared, Take cube roots.; equals 2 x squared y squared the cube root of 3 y squared")[$root(3, 24 x^(6) y^(8)) & = root(3, 8 x^(6) y^(6) ⋅ 3 y^(2)) &  & "Apply Property (1)." \  & = root(3, 8 x^(6) y^(6)) root(3, 3 y^(2)) &  & "Take cube roots." \  & = 2 x^(2) y^(2) root(3, 3 y^(2))$]
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Property (1) applies only to products under the radical, not to sums or differences. Thus, for example,


        #math.equation(block: true, alt: "the square root of 4 times 9 equals the square root of 4 the square root of 9 equals 2 times 3 , but the square root of 4 plus 9 not equal to the square root of 4 plus the square root of 9")[$sqrt(4 ⋅ 9) = sqrt(4) sqrt(9) = 2 ⋅ 3 , " " " " " " " but " " " " " " " sqrt(4 + 9) ≠ sqrt(4) + sqrt(9)$]
        and


        #math.equation(block: true, alt: "the cube root of x cubed y to the power 6 equals the cube root of x cubed the cube root of y to the power 6 equals x y squared , but the cube root of x cubed minus y to the power 6 not equal to the cube root of x cubed minus the cube root of y to the power 6")[$root(3, x^(3) y^(6)) = root(3, x^(3)) root(3, y^(6)) = x y^(2) , " " " " " " " but " " " " " " " root(3, x^(3) − y^(6)) ≠ root(3, x^(3)) − root(3, y^(6))$]
]


      To simplify roots of fractions, we use Property (2), which allows us to write the expression as a quotient of two radicals.


      #examplebox("Example 3")[][
+ #math.equation(block: false, alt: "the square root of the fraction 3 over 4 equals the fraction the square root of 3 over the square root of 4 equals the fraction the square root of 3 over 2")[$display(sqrt(frac(3, 4)) = frac(sqrt(3), sqrt(4)) = frac(sqrt(3), 2))$]
+ #math.equation(block: false, alt: "the cube root of the fraction 5 over 8 equals the fraction the cube root of 5 over the cube root of 8 equals the fraction the cube root of 5 over 2")[$display(root(3, frac(5, 8)) = frac(root(3, 5), root(3, 8)) = frac(root(3, 5), 2))$]
]


      We can also use Properties (1) and (2) to simplify products and quotients of radicals.


      #examplebox("Example 4")[][
Simplify.


          + #math.equation(block: false, alt: "the 4th root of 6 x squared the 4th root of 8 x cubed")[$root(4, 6 x^(2)) root(4, 8 x^(3))$]
+ #math.equation(block: false, alt: "the fraction the cube root of 16 y to the power 5 over the cube root of y squared")[$display(frac(root(3, 16 y^(5)), root(3, y^(2))))$]


        
        #solutionbox[
+ First apply Property (1) to write the product as a single radical, then simplify.
            #math.equation(block: true, alt: "the 4th root of 6 x squared the 4th root of 8 x cubed, equals the 4th root of 48 x to the power 5, Factor out perfect fourth powers.; equals the 4th root of 16 x to the power 4 the 4th root of 3 x, Simplify.; equals 2 x the 4th root of 3 x")[$root(4, 6 x^(2)) root(4, 8 x^(3)) & = root(4, 48 x^(5)) &  & "Factor out perfect fourth powers." \  & = root(4, 16 x^(4)) root(4, 3 x) &  & "Simplify." \  & = 2 x root(4, 3 x)$]
+ Apply Property (2) to write the quotient as a single radical.
            #math.equation(block: true, alt: "the fraction the cube root of 16 y to the power 5 over the cube root of y squared, equals the cube root of the fraction 16 y to the power 5 over y squared, Reduce.; equals the cube root of 16 y cubed, Simplify: factor out perfect cubes.; equals the cube root of 8 y cubed the cube root of 2; equals 2 y the cube root of 2")[$frac(root(3, 16 y^(5)), root(3, y^(2))) & = root(3, frac(16 y^(5), y^(2))) &  & "Reduce." \  & = root(3, 16 y^(3)) &  & "Simplify: factor out perfect cubes." \  & = root(3, 8 y^(3)) root(3, 2) \  & = 2 y root(3, 2)$]
]
]


    
    
      
=== Sums and Differences of Radicals


      You know that sums or differences of like terms can be combined by adding or subtracting their coefficients:


      #math.equation(block: true, alt: "3 x y plus 5 x y equals open parenthesis 3 plus 5 close parenthesis x y equals 8 x y")[$3 x y + 5 x y = ( 3 + 5 ) x y = 8 x y$]
      Like radicals, that is, radicals of the same index and radicand, can be combined in the same way.


      #examplebox("Example 5")[][
+ #math.equation(block: false, alt: "3 the square root of 3 plus 4 the square root of 3, equals open parenthesis 3 plus 4 close parenthesis the square root of 3; equals 7 the square root of 3")[$3 sqrt(3) + 4 sqrt(3) & = ( 3 + 4 ) sqrt(3) \  & = 7 sqrt(3)$]
+ #math.equation(block: false, alt: "4 the cube root of y minus 6 the cube root of y, equals open parenthesis 4 minus 6 close parenthesis the cube root of y; equals minus 2 the cube root of y")[$4 root(3, y) − 6 root(3, y) & = ( 4 − 6 ) root(3, y) \  & = − 2 root(3, y)$]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ In Examplea, #math.equation(block: false, alt: "3 the square root of 3 plus 4 the square root of 3 not equal to 7 the square root of 6")[$3 sqrt(3) + 4 sqrt(3) ≠ 7 sqrt(6)$]. Only the coefficients are added; the radicand does not change.
+ Sums of radicals with different radicands or different indices cannot be combined. Thus, 
                #math.equation(block: true, alt: "the square root of 11 plus the square root of 5 not equal to the square root of 16")[$sqrt(11) + sqrt(5) ≠ sqrt(16)$]
                #math.equation(block: true, alt: "the cube root of 10 x minus the cube root of 2 x not equal to the cube root of 8 x")[$root(3, 10 x) − root(3, 2 x) ≠ root(3, 8 x)$]
            and
                #math.equation(block: true, alt: "the cube root of 7 plus the square root of 7 not equal to the 5th root of 7")[$root(3, 7) + sqrt(7) ≠ root(5, 7)$]
            None of the expressions above can be simplified.
]


    
    
      
=== Products of Radicals


      According to Property (1), radicals of the same index can be multiplied together.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Product of Radicals]


        #math.equation(block: true, alt: "the nth root of a the nth root of b equals the nth root of a b open parenthesis a , b greater than or equal to 0 close parenthesis")[$root(n, a) root(n, b) = root(n, a b) #hide($b l a n k b l a n k$) ( a , b ≥ 0 )$]
]


      Thus, for example,


      #math.equation(block: true, alt: "the square root of 2 the square root of 18 the square root of 36 equals 6 and the cube root of 2 x the cube root of 4 x squared equals the cube root of 8 x cubed equals 2 x")[$sqrt(2) sqrt(18) sqrt(36) = 6 " " " " " " " and " " " " " " " root(3, 2 x) root(3, 4 x^(2)) = root(3, 8 x^(3)) = 2 x$]
      For products involving binomials, we can apply the distributive law.


      #examplebox("Example 6")[][
+ #math.equation(block: false, alt: "the square root of 3 open parenthesis the square root of 2 x plus the square root of 6 close parenthesis, equals the square root of 3 times 2 x plus the square root of 3 times 6; equals the square root of 6 x plus the square root of 18 equals the square root of 6 x plus 3 the square root of 2")[$sqrt(3) ( sqrt(2 x) + sqrt(6) ) & = sqrt(3 ⋅ 2 x) + sqrt(3 ⋅ 6) \  & = sqrt(6 x) + sqrt(18) = sqrt(6 x) + 3 sqrt(2)$]
+ #math.equation(block: false, alt: "open parenthesis the square root of x minus the square root of y close parenthesis open parenthesis the square root of x plus the square root of y close parenthesis, equals the square root of x squared plus the square root of x y minus the square root of x y minus the square root of y squared; equals x minus y")[$( sqrt(x) − sqrt(y) ) ( sqrt(x) + sqrt(y) ) & = sqrt(x^(2)) + sqrt(x y) − sqrt(x y) − sqrt(y^(2)) \  & = x − y$]
]


    
    
      
=== Rationalizing the Denominator


      It is easier to work with radicals if there are no roots in the denominators of fractions. We can use the fundamental principle of fractions to remove radicals from the denominator. This process is called #strong[rationalizing the denominator]. For square roots, we multiply the numerator and denominator of the fraction by the radical in the denominator.


      #examplebox("Example 7")[][
Rationalize the denominator of each fraction.


          + #math.equation(block: false, alt: "the square root of the fraction 1 over 3")[$display(sqrt(frac(1, 3)))$]
+ #math.equation(block: false, alt: "the fraction the square root of 2 over the square root of 50 x")[$display(frac(sqrt(2), sqrt(50 x)))$]


        
        #solutionbox[
+ Apply Property (2) to write the radical as a quotient. #linebreak() #math.equation(block: false, alt: "the square root of the fraction 1 over 3, equals the fraction the square root of 1 over the square root of 3; equals the fraction 1 over the square root of 3, Multiply numerator and denominator by the square root of 3.; equals the fraction 1 times the square root of 3 over the square root of 3 times the square root of 3; equals the fraction the square root of 3 over 3")[$sqrt(frac(1, 3)) & = frac(sqrt(1), sqrt(3)) \  & = frac(1, sqrt(3)) &  & "Multiply numerator and denominator by " sqrt(3) . \  & = frac(1 ⋅ sqrt(3), sqrt(3) ⋅ sqrt(3)) \  & = frac(sqrt(3), 3)$]
+ It is always best to simplify the denominator before rationalizing. #linebreak() #math.equation(block: false, alt: "the fraction the square root of 2 over the square root of 50 x, equals the fraction the square root of 2 over 5 the square root of 2 x, Multiply numerator and denominator by the square root of 2 x .; equals the fraction the square root of 2 times the square root of 2 x over 5 the square root of 2 x times the square root of 2 x, Simplify.; equals the fraction the square root of 4 x over 5 open parenthesis 2 x close parenthesis; equals the fraction 2 the square root of x over 10 x equals the fraction the square root of x over 5 x")[$frac(sqrt(2), sqrt(50 x)) & = frac(sqrt(2), 5 sqrt(2 x)) &  & "Multiply numerator and denominator by " sqrt(2 bold(italic(x))) . \  & = frac(sqrt(2) ⋅ sqrt(2 bold(italic(x))), 5 sqrt(2 x) ⋅ sqrt(2 bold(italic(x)))) &  & "Simplify." \  & = frac(sqrt(4 x), 5 ( 2 x )) \  & = frac(2 sqrt(x), 10 x) = frac(sqrt(x), 5 x)$]
]
]


      If the denominator of a fraction is a #emph[binomial] in which one or both terms is a radical, we can use a special building factor to rationalize it. First, recall that


      #math.equation(block: true, alt: "open parenthesis p minus q close parenthesis open parenthesis p plus q close parenthesis equals p squared minus q squared")[$( p − q ) ( p + q ) = p^(2) − q^(2)$]
      where the product consists of perfect squares only. Each of the two factors #math.equation(block: false, alt: "p minus q")[$p − q$] and #math.equation(block: false, alt: "p plus q")[$p + q$] is said to be the #strong[conjugate] of the other.


      Now consider a fraction of the form


      #math.equation(block: true, alt: "the fraction a over b plus the square root of c")[$frac(a, b + sqrt(c))$]
      If we multiply the numerator and denominator of this fraction by the conjugate of the denominator, we get


      #math.equation(block: true, alt: "the fraction a open parenthesis b minus the square root of c close parenthesis over open parenthesis b plus the square root of c close parenthesis open parenthesis b minus the square root of c close parenthesis equals the fraction a b minus a the square root of c over b squared minus open parenthesis the square root of c close parenthesis squared equals the fraction a b minus a the square root of c over b squared minus c")[$frac(a ( bold(italic(b)) − sqrt(bold(italic(c))) ), ( b + sqrt(c) ) ( bold(italic(b)) − sqrt(bold(italic(c))) )) = frac(a b − a sqrt(c), b^(2) − ( sqrt(c) )^(2)) = frac(a b − a sqrt(c), b^(2) − c)$]
      The denominator of the fraction no longer contains any radicals—it has been rationalized.


      Multiplying numerator and denominator by the conjugate of the denominator also works on fractions of the form


      #math.equation(block: true, alt: "the fraction a over the square root of b plus c and the fraction a over the square root of b plus the square root of c")[$frac(a, sqrt(b) + c) " " " " " " " and " " " " " " " frac(a, sqrt(b) + sqrt(c))$]
      We leave the verification of these cases as exercises.


      #examplebox("Example 8")[][
Rationalize the denominator: #math.equation(block: false, alt: "the fraction x over the square root of 2 plus the square root of x")[$display(frac(x, sqrt(2) + sqrt(x)))$].


        
        #solutionbox[
Multiply numerator and denominator by the conjugate of the denominator, #math.equation(block: false, alt: "the square root of 2 minus the square root of x")[$sqrt(2) − sqrt(x)$].


          #math.equation(block: true, alt: "the fraction x open parenthesis the square root of 2 minus the square root of x close parenthesis over open parenthesis the square root of 2 plus the square root of x close parenthesis open parenthesis the square root of 2 minus the square root of x close parenthesis equals the fraction x open parenthesis the square root of 2 minus the square root of x close parenthesis over 2 minus x")[$frac(x ( sqrt(2) − sqrt(bold(italic(x))) ), ( sqrt(2) + sqrt(x) ) ( sqrt(2) − sqrt(bold(italic(x))) )) = frac(x ( sqrt(2) − sqrt(x) ), 2 − x)$]
]
]


    
    
      
=== Simplifying #math.equation(block: false, alt: "the nth root of x to the power n")[$root(n, x^(n))$]


      Raising to a power is the inverse operation for extracting roots; that is,


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "open parenthesis the nth root of a close parenthesis to the power n equals a")[$attach(( root(n, a) ), t: n) = a$]
]


      as long as #math.equation(block: false, alt: "the nth root of a")[$root(n, a)$] is a real number. For example,


      #math.equation(block: true, alt: "open parenthesis the 4th root of 16 close parenthesis to the power 4 equals 2 to the power 4 equals 16 , and open parenthesis the cube root of minus 125 close parenthesis cubed equals open parenthesis minus 5 close parenthesis cubed equals minus 125")[$attach(( root(4, 16) ), t: 4) = 2^(4) = 16 , " " " " " " " and " " " " " " " attach(( root(3, − 125) ), t: 3) = ( − 5 )^(3) = − 125$]
      Now consider the power and root operations in the opposite order; is it true that #math.equation(block: false, alt: "the nth root of a to the power n equals a")[$root(n, a^(n)) = a$]? If the index #math.equation(block: false, alt: "n")[$n$] is an odd number, then the statement is always true. For example,


      #math.equation(block: true, alt: "the cube root of 2 cubed equals the cube root of 8 equals 2 and the cube root of open parenthesis minus 2 close parenthesis cubed equals the cube root of minus 8 equals minus 2")[$root(3, 2^(3)) = root(3, 8) = 2 " " " " " " " and " " " " " " " root(3, ( − 2 )^(3)) = root(3, − 8) = − 2$]
      However, if #math.equation(block: false, alt: "n")[$n$] is even, we must be careful. Recall that the principal root #math.equation(block: false, alt: "the nth root of x")[$root(n, x)$] is always positive, so if #math.equation(block: false, alt: "a")[$a$] is a negative number, it cannot be true that #math.equation(block: false, alt: "the nth root of a to the power n equals a")[$root(n, a^(n)) = a$]. For example, if #math.equation(block: false, alt: "a equals minus 3")[$a = − 3$], then


      #math.equation(block: true, alt: "the square root of open parenthesis minus 3 close parenthesis squared equals the square root of 9 equals 3")[$sqrt(( − 3 )^(2)) = sqrt(9) = 3$]
      Instead, we see that, for even roots, #math.equation(block: false, alt: "the nth root of a to the power n equals vertical bar a vertical bar")[$root(n, a^(n)) = | a |$].


      We summarize our results in below.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Roots of Powers]


        + If #math.equation(block: false, alt: "n")[$n$] is odd, #math.equation(block: false, alt: "the nth root of a to the power n equals a")[$#hide($b l a n k b l a n k$) root(n, a^(n)) = a$]
+ If #math.equation(block: false, alt: "n")[$n$] is even, #math.equation(block: false, alt: "the nth root of a to the power n equals vertical bar a vertical bar")[$#hide($b l a n k b l a n k$) root(n, a^(n)) = | a |$] #linebreak() In particular, #math.equation(block: false, alt: "the square root of a squared equals vertical bar a vertical bar")[$#hide($b l a n k b l a n k$) sqrt(a^(2)) = | a |$]
]


      #examplebox("Example 9")[][
+ #math.equation(block: false, alt: "the square root of 16 x squared equals 4 vertical bar x vertical bar")[$sqrt(16 x^(2)) = 4 | x |$]
+ #math.equation(block: false, alt: "the square root of open parenthesis x minus 1 close parenthesis squared equals vertical bar x minus 1 vertical bar")[$sqrt(( x − 1 )^(2)) = | x − 1 |$]
]


    
    
      
=== Extraneous Solutions to Radical Equations


      It is important to check the solution to a radical equation, because it is possible to introduce false, or #strong[extraneous], solutions when we square both sides of the equation. For example, the equation


      #math.equation(block: true, alt: "the square root of x equals minus 5")[$sqrt(x) = − 5$]
      has no solution, because #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$] is never a negative number. However, if we try to solve the equation by squaring both sides, we find


      #math.equation(block: true, alt: "open parenthesis the square root of x close parenthesis squared, equals open parenthesis minus 5 close parenthesis squared; x, equals 25")[$( sqrt(x) )^(2) & = ( − 5 )^(2) \ x & = 25$]
      You can check that #math.equation(block: false, alt: "25")[$25$] is #emph[not] a solution to the original equation, #math.equation(block: false, alt: "the square root of x equals minus 5")[$sqrt(x) = − 5$], because #math.equation(block: false, alt: "the square root of 25")[$sqrt(25)$] does not equal #math.equation(block: false, alt: "minus 5")[$− 5$].


      If each side of an equation is raised to an odd power, extraneous solutions will not be introduced. However, if we raise both sides to an even power, we should check each solution in the original equation.


      #examplebox("Example 10")[][
Solve the equation #math.equation(block: false, alt: "the square root of x plus 2 plus 4 equals x")[$sqrt(x + 2) + 4 = x$].


        
        #solutionbox[
First, isolate the radical expression on one side of the equation. (This will make it easier to square both sides.)


          #math.equation(block: true, alt: "the square root of x plus 2, equals x minus 4, Square both sides of the equation.; open parenthesis the square root of x plus 2 close parenthesis squared, equals open parenthesis x minus 4 close parenthesis squared; x plus 2, equals x squared minus 8 x plus 16, Subtract x plus 2 from both sides.; x squared minus 9 x plus 14, equals 0, Factor the left side.; x equals 2, or x equals 7")[$sqrt(x + 2) & = x − 4 &  & "Square both sides of the equation." \ attach(( sqrt(x + 2) ), t: 2) & = ( x − 4 )^(2) \ x + 2 & = x^(2) − 8 x + 16 &  & "Subtract " bold(italic(x)) + 2 " from both sides." \ x^(2) − 9 x + 14 & = 0 &  & "Factor the left side." \ x = 2 #hide($000$) & "or" #hide($000$) x = 7$]
          #emph[Check]


          Does #math.equation(block: false, alt: "the square root of 2 plus 2 plus 4 equals 2")[$sqrt(2 + 2) + 4 = 2$]?


          #math.equation(block: true, alt: "No; 2 is not a solution.")[$#hide($b l a n k$) "No; " 2 " is not a solution."$]


          Does #math.equation(block: false, alt: "the square root of 7 plus 2 plus 4 equals 7")[$sqrt(7 + 2) + 4 = 7$]?


          #math.equation(block: true, alt: "Yes; 7 is a solution.")[$#hide($b l a n k$) "Yes; " 7 " is a solution."$]


          The apparent solution #math.equation(block: false, alt: "2")[$2$] is extraneous. The only solution to the original equation is #math.equation(block: false, alt: "7")[$7$]. We can verify the solution by graphing the equations


          #math.equation(block: true, alt: "y sub 1 equals the square root of x plus 2 and y sub 2 equals x minus 4")[$y_(1) = sqrt(x + 2) " " " " " " " and " " " " " " " y_(2) = x − 4$]
          as shown at right. The graphs intersect in only one point, #math.equation(block: false, alt: "open parenthesis 7 , 3 close parenthesis")[$( 7 , 3 )$], so there is only one solution, #math.equation(block: false, alt: "x equals 7")[$x = 7$].


          #figure(figph[graphs of both sides of radical equation], alt: "graphs of both sides of radical equation", caption: none)
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
When we square both sides of an equation, it is #emph[not] correct to square each term of the equation separately. Thus, in Example, the original equation is not equivalent to


        #math.equation(block: true, alt: "open parenthesis the square root of x plus 2 close parenthesis squared plus 4 squared equals x squared")[$( sqrt(x + 2) )^(2) + 4^(2) = x^(2)$]
        This is because #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared not equal to a squared plus b squared")[$( a + b )^(2) ≠ a^(2) + b^(2)$]. Instead, we must square the #emph[entire] left side of the equation as a binomial, like this,


        #math.equation(block: true, alt: "open parenthesis the square root of x plus 2 plus 4 close parenthesis squared equals x squared")[$( sqrt(x + 2) + 4 )^(2) = x^(2)$]
        or we may proceed as shown in Example.
]


    
    
      
=== Equations with More than One Radical


      Sometimes it is necessary to square both sides of an equation more than once in order to eliminate all the radicals.


      #examplebox("Example 11")[][
Solve #math.equation(block: false, alt: "the square root of x minus 7 plus the square root of x equals 7")[$" " sqrt(x − 7) + sqrt(x) = 7$].


        
        #solutionbox[
First, isolate the more complicated radical on one side of the equation. (This will make it easier to square both sides.) We will subtract #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$] from both sides.


          #math.equation(block: true, alt: "the square root of x minus 7 equals 7 minus the square root of x")[$sqrt(x − 7) = 7 − sqrt(x)$]
          Now square each side to remove one radical. Be careful when squaring the binomial #math.equation(block: false, alt: "7 minus the square root of x")[$7 − sqrt(x)$].


          #math.equation(block: true, alt: "open parenthesis the square root of x minus 7 close parenthesis squared, equals open parenthesis 7 minus the square root of x close parenthesis squared; x minus 7, equals 49 minus 14 the square root of x plus x")[$( sqrt(x − 7) )^(2) & = ( 7 − sqrt(x) )^(2) \ x − 7 & = 49 − 14 sqrt(x) + x$]
          Collect like terms, and isolate the radical on one side of the equation.


          #math.equation(block: true, alt: "minus 56, equals minus 14 the square root of x, Divide both sides by minus 14.; 4, equals the square root of x")[$− 56 & = − 14 sqrt(x) &  & "Divide both sides by " − 14. \ 4 & = sqrt(x)$]
          Now square again to obtain


          #math.equation(block: true, alt: "open parenthesis 4 close parenthesis squared, equals open parenthesis the square root of x close parenthesis squared; 16, equals x")[$( 4 )^(2) & = ( sqrt(x) )^(2) \ 16 & = x$]
          #emph[Check]


          Does #math.equation(block: false, alt: "the square root of 16 minus 7 plus the square root of 16 equals 7")[$sqrt(16 − 7) + sqrt(16) = 7$]?


          #math.equation(block: true, alt: "Yes. The solution is 16.")[$#hide($b l a n k$) "Yes. The solution is " 16.$]
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Recall that we cannot solve a radical equation by squaring each term separately. In other words, it is #emph[incorrect] to begin Example by writing


        #math.equation(block: true, alt: "open parenthesis the square root of x minus 7 close parenthesis squared plus open parenthesis the square root of x close parenthesis squared equals 7 squared")[$( sqrt(x − 7) )^(2) + ( sqrt(x) )^(2) = 7^(2)$]
        We must square the #emph[entire expression] on each side of the equal sign as one piece.
]


    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Radical
- Extraneous
- Rationalize
- Conjugate
- Like radicals
- Radicand
- Index


      
      
        
==== SKILLS


        Practice each skill in the exercises listed.


        + Simplify radicals: \#1–6
+ Simplify products and quotients of radcials: \#7–10
+ Combine like radicals: \#11–18
+ Multiply radical expressions: \#19–36
+ Rationalize the denominator: \#37–50
+ Simplify #math.equation(block: false, alt: "the nth root of a to the power n")[$root(n, a^(n))$]: \#51–54
+ Solve radical equations: \#55–80


      
    
    
      
=== Exercises A.10


      For Problems 1-6, simplify. Assume that all variables represent positive numbers.


      
        + #math.equation(block: false, alt: "the square root of 18")[$sqrt(18)$]
+ #math.equation(block: false, alt: "the cube root of 24")[$root(3, 24)$]
+ #math.equation(block: false, alt: "minus the 4th root of 64")[$− root(4, 64)$]


        
          + #math.equation(block: false, alt: "3 the square root of 2")[$3 sqrt(2)$]
+ #math.equation(block: false, alt: "2 the cube root of 3")[$2 root(3, 3)$]
+ #math.equation(block: false, alt: "minus 2 the 4th root of 4 equals minus 2 the square root of 2")[$− 2 root(4, 4) = − 2 sqrt(2)$]


        
      
      
        + #math.equation(block: false, alt: "the square root of 50")[$sqrt(50)$]
+ #math.equation(block: false, alt: "the cube root of 54")[$root(3, 54)$]
+ #math.equation(block: false, alt: "minus the 4th root of 162")[$− root(4, 162)$]


      
      
        + #math.equation(block: false, alt: "the square root of 60 , 000")[$sqrt(60 "," 000)$]
+ #math.equation(block: false, alt: "the cube root of 900 , 000")[$root(3, 900 "," 000)$]
+ #math.equation(block: false, alt: "the cube root of the fraction minus 40 over 27")[$root(3, display(frac(− 40, 27)))$]


        
          + #math.equation(block: false, alt: "100 the square root of 6")[$100 sqrt(6)$]
+ #math.equation(block: false, alt: "10 the cube root of 900")[$10 root(3, 900)$]
+ #math.equation(block: false, alt: "the fraction minus 2 over 3 the cube root of 5")[$display(frac(− 2, 3)) root(3, 5)$]


        
      
      
        + #math.equation(block: false, alt: "the square root of 800 , 000")[$sqrt(800 "," 000)$]
+ #math.equation(block: false, alt: "the cube root of 24 , 000")[$root(3, 24 "," 000)$]
+ #math.equation(block: false, alt: "the 4th root of the fraction 80 over 625")[$root(4, display(frac(80, 625)))$]


      
      
        + #math.equation(block: false, alt: "the cube root of x to the power 10")[$root(3, x^(10))$]
+ #math.equation(block: false, alt: "the square root of 27 z cubed")[$sqrt(27 z^(3))$]
+ #math.equation(block: false, alt: "the 4th root of 48 a to the power 9")[$root(4, 48 a^(9))$]


        
          + #math.equation(block: false, alt: "x cubed the cube root of x")[$x^(3) root(3, x)$]
+ #math.equation(block: false, alt: "3 z the square root of 3 z")[$3 z sqrt(3 z)$]
+ #math.equation(block: false, alt: "2 a squared the 4th root of 3 a")[$2 a^(2) root(4, 3 a)$]


        
      
      
        + #math.equation(block: false, alt: "the cube root of y to the power 16")[$root(3, y^(16))$]
+ #math.equation(block: false, alt: "the square root of 12 t to the power 5")[$sqrt(12 t^(5))$]
+ #math.equation(block: false, alt: "the cube root of 81 b to the power 8")[$root(3, 81 b^(8))$]


      
      For Problems 7-10, simplify.


      
        + #math.equation(block: false, alt: "minus the square root of 18 s the square root of 2 s cubed")[$− sqrt(18 s) sqrt(2 s^(3))$]
+ #math.equation(block: false, alt: "the cube root of 7 h squared the cube root of minus 49 h")[$root(3, 7 h^(2)) root(3, − 49 h)$]
+ #math.equation(block: false, alt: "the square root of 16 minus 4 x squared")[$sqrt(16 − 4 x^(2))$]


        
          + #math.equation(block: false, alt: "minus 6 s squared")[$− 6 s^(2)$]
+ #math.equation(block: false, alt: "minus 7 h")[$− 7 h$]
+ #math.equation(block: false, alt: "2 the square root of 4 minus x squared")[$2 sqrt(4 − x^(2))$]


        
      
      
        + #math.equation(block: false, alt: "the square root of 3 w cubed the square root of 27 w cubed")[$sqrt(3 w^(3)) sqrt(27 w^(3))$]
+ #math.equation(block: false, alt: "minus the 4th root of 2 m cubed the 4th root of 8 m")[$− root(4, 2 m^(3)) root(4, 8 m)$]
+ #math.equation(block: false, alt: "the square root of 9 Y squared plus 18")[$sqrt(9 Y^(2) + 18)$]


      
      
        + #math.equation(block: false, alt: "the cube root of 8 A cubed plus A to the power 6")[$root(3, 8 A^(3) + A^(6))$]
+ #math.equation(block: false, alt: "the fraction the square root of 45 x cubed y cubed over the square root of 5 y")[$display(frac(sqrt(45 x^(3) y^(3)), sqrt(5 y)))$]
+ #math.equation(block: false, alt: "the fraction the cube root of 8 b to the power 7 over the cube root of a to the power 6 b squared")[$display(frac(root(3, 8 b^(7)), root(3, a^(6) b^(2))))$]


        
          + #math.equation(block: false, alt: "A the cube root of 8 plus A cubed")[$A root(3, 8 + A^(3))$]
+ #math.equation(block: false, alt: "3 x y the square root of x")[$3 x y sqrt(x)$]
+ #math.equation(block: false, alt: "the fraction 2 b the cube root of b squared over a squared")[$display(frac(2 b root(3, b^(2)), a^(2)))$]


        
      
      
        + #math.equation(block: false, alt: "the cube root of b to the power 9 minus 27 b cubed")[$root(3, b^(9) − 27 b^(3))$]
+ #math.equation(block: false, alt: "the fraction the square root of 98 x squared y cubed over the square root of x y")[$display(frac(sqrt(98 x^(2) y^(3)), sqrt(x y)))$]
+ #math.equation(block: false, alt: "the fraction the cube root of 16 r to the power 4 over the cube root of 4 t cubed")[$display(frac(root(3, 16 r^(4)), root(3, 4 t^(3))))$]


      
      For Problems 11-18, simplify and combine like terms.


      
        #math.equation(block: true, alt: "3 the square root of 7 plus 2 the square root of 7")[$3 sqrt(7) + 2 sqrt(7)$]


        
          #math.equation(block: true, alt: "5 the square root of 7")[$5 sqrt(7)$]


        
      
      
        #math.equation(block: true, alt: "5 the square root of 2 minus 3 the square root of 2")[$5 sqrt(2) − 3 sqrt(2)$]


      
      
        #math.equation(block: true, alt: "4 the square root of 3 minus the square root of 27")[$4 sqrt(3) − sqrt(27)$]


        
          #math.equation(block: true, alt: "the square root of 3")[$sqrt(3)$]


        
      
      
        #math.equation(block: true, alt: "the square root of 75 plus 2 the square root of 3")[$sqrt(75) + 2 sqrt(3)$]


      
      
        #math.equation(block: true, alt: "the square root of 50 x plus the square root of 32 x")[$sqrt(50 x) + sqrt(32 x)$]


        
          #math.equation(block: true, alt: "9 the square root of 2 x")[$9 sqrt(2 x)$]


        
      
      
        #math.equation(block: true, alt: "the square root of 8 y minus the square root of 18 y")[$sqrt(8 y) − sqrt(18 y)$]


      
      
        #math.equation(block: true, alt: "3 the cube root of 16 minus the cube root of 2 minus 2 the cube root of 54")[$3 root(3, 16) − root(3, 2) − 2 root(3, 54)$]


        
          #math.equation(block: true, alt: "minus the cube root of 2")[$− root(3, 2)$]


        
      
      
        #math.equation(block: true, alt: "the cube root of 81 plus 2 the cube root of 24 minus 3 the cube root of 3")[$root(3, 81) + 2 root(3, 24) − 3 root(3, 3)$]


      
      For Problems 19-32, multiply.


      
        #math.equation(block: true, alt: "2 open parenthesis 3 minus the square root of 5 close parenthesis")[$2 ( 3 − sqrt(5) )$]


        
          #math.equation(block: true, alt: "6 minus 2 the square root of 5")[$6 − 2 sqrt(5)$]


        
      
      
        #math.equation(block: true, alt: "5 open parenthesis 2 minus the square root of 7 close parenthesis")[$5 ( 2 − sqrt(7) )$]


      
      
        #math.equation(block: true, alt: "the square root of 2 open parenthesis the square root of 6 plus the square root of 10 close parenthesis")[$sqrt(2) ( sqrt(6) + sqrt(10) )$]


        
          #math.equation(block: true, alt: "2 the square root of 3 plus 2 the square root of 5")[$2 sqrt(3) + 2 sqrt(5)$]


        
      
      
        #math.equation(block: true, alt: "the square root of 3 open parenthesis the square root of 12 minus the square root of 15 close parenthesis")[$sqrt(3) ( sqrt(12) − sqrt(15) )$]


      
      
        #math.equation(block: true, alt: "the cube root of 2 open parenthesis the cube root of 20 minus 2 the cube root of 12 close parenthesis")[$root(3, 2) ( root(3, 20) − 2 root(3, 12) )$]


        
          #math.equation(block: true, alt: "2 the cube root of 5 minus 4 the cube root of 3")[$2 root(3, 5) − 4 root(3, 3)$]


        
      
      
        #math.equation(block: true, alt: "the cube root of 3 open parenthesis 2 the cube root of 18 plus the cube root of 36 close parenthesis")[$root(3, 3) ( 2 root(3, 18) + root(3, 36) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis the square root of x minus 3 close parenthesis open parenthesis the square root of x plus 3 close parenthesis")[$( sqrt(x) − 3 ) ( sqrt(x) + 3 )$]


        
          #math.equation(block: true, alt: "x minus 9")[$x − 9$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 2 plus the square root of x close parenthesis open parenthesis 2 minus the square root of x close parenthesis")[$( 2 + sqrt(x) ) ( 2 − sqrt(x) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 2 minus the square root of 3 close parenthesis open parenthesis the square root of 2 plus 2 the square root of 3 close parenthesis")[$( sqrt(2) − sqrt(3) ) ( sqrt(2) + 2 sqrt(3) )$]


        
          #math.equation(block: true, alt: "minus 4 plus the square root of 6")[$− 4 + sqrt(6)$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 3 minus the square root of 5 close parenthesis open parenthesis 2 the square root of 3 plus the square root of 5 close parenthesis")[$( sqrt(3) − sqrt(5) ) ( 2 sqrt(3) + sqrt(5) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 5 minus the square root of 2 close parenthesis squared")[$( sqrt(5) − sqrt(2) )^(2)$]


        
          #math.equation(block: true, alt: "7 minus 2 the square root of 10")[$7 − 2 sqrt(10)$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 2 minus 2 the square root of 3 close parenthesis squared")[$( sqrt(2) − 2 sqrt(3) )^(2)$]


      
      
        #math.equation(block: true, alt: "open parenthesis the square root of a minus 2 the square root of b close parenthesis squared")[$( sqrt(a) − 2 sqrt(b) )^(2)$]


        
          #math.equation(block: true, alt: "a minus 4 the square root of a b plus 4 b")[$a − 4 sqrt(a b) + 4 b$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 2 a minus 2 the square root of b close parenthesis open parenthesis the square root of 2 a plus 2 the square root of b close parenthesis")[$( sqrt(2 a) − 2 sqrt(b) ) ( sqrt(2 a) + 2 sqrt(b) )$]


      
      For Problems 33-36, verify by substitution that the number is a solution of the quadratic equation.


      
        #math.equation(block: true, alt: "x squared minus 2 x minus 2 equals 0 , 1 plus the square root of 3")[$x^(2) − 2 x − 2 = 0 ", " " " 1 + sqrt(3)$]


        
          #math.equation(block: true, alt: "open parenthesis 1 plus the square root of 3 close parenthesis squared minus 2 open parenthesis 1 plus the square root of 3 close parenthesis minus 2 equals 0")[$( 1 + sqrt(3) )^(2) − 2 ( 1 + sqrt(3) ) − 2 = 0$]


        
      
      
        #math.equation(block: true, alt: "x squared plus 4 x minus 1 equals 0 , minus 2 plus the square root of 5")[$x^(2) + 4 x − 1 = 0 ", " " " − 2 + sqrt(5)$]


      
      
        #math.equation(block: true, alt: "x squared plus 6 x minus 9 equals 0 , minus 3 plus 3 the square root of 2")[$x^(2) + 6 x − 9 = 0 ", " " " − 3 + 3 sqrt(2)$]


        
          #math.equation(block: true, alt: "open parenthesis minus 3 plus 3 the square root of 2 close parenthesis squared plus 6 open parenthesis minus 3 plus 3 the square root of 2 close parenthesis minus 9 equals 0")[$( − 3 + 3 sqrt(2) )^(2) + 6 ( − 3 + 3 sqrt(2) ) − 9 = 0$]


        
      
      
        #math.equation(block: true, alt: "4 x squared minus 20 x plus 22 equals 0 , the fraction 5 minus the square root of 3 over 2")[$4 x^(2) − 20 x + 22 = 0 ", " " " display(frac(5 − sqrt(3), 2))$]


      
      For Problems 37-50, rationalize the denominator.


      
        #math.equation(block: true, alt: "the fraction 6 over the square root of 3")[$display(frac(6, sqrt(3)))$]


        
          #math.equation(block: true, alt: "2 the square root of 3")[$2 sqrt(3)$]


        
      
      
        #math.equation(block: true, alt: "the fraction 10 over the square root of 5")[$display(frac(10, sqrt(5)))$]


      
      
        #math.equation(block: true, alt: "the square root of the fraction 7 x over 18")[$sqrt(display(frac(7 x, 18)))$]


        
          #math.equation(block: true, alt: "the fraction the square root of 14 x over 6")[$display(frac(sqrt(14 x), 6))$]


        
      
      
        #math.equation(block: true, alt: "the square root of the fraction 27 x over 20")[$sqrt(display(frac(27 x, 20)))$]


      
      
        #math.equation(block: true, alt: "the square root of the fraction 2 a over b")[$sqrt(display(frac(2 a, b)))$]


        
          #math.equation(block: true, alt: "the fraction the square root of 2 a b over b")[$display(frac(sqrt(2 a b), b))$]


        
      
      
        #math.equation(block: true, alt: "the square root of the fraction 5 p over q")[$sqrt(display(frac(5 p, q)))$]


      
      
        #math.equation(block: true, alt: "the fraction 2 the square root of 3 over the square root of 2 k")[$display(frac(2 sqrt(3), sqrt(2 k)))$]


        
          #math.equation(block: true, alt: "the fraction the square root of 6 k over k")[$display(frac(sqrt(6 k), k))$]


        
      
      
        #math.equation(block: true, alt: "the fraction 6 the square root of 2 over the square root of 3 v")[$display(frac(6 sqrt(2), sqrt(3 v)))$]


      
      
        #math.equation(block: true, alt: "the fraction 4 over 1 plus the square root of 3")[$display(frac(4, 1 + sqrt(3)))$]


        
          #math.equation(block: true, alt: "minus 2 open parenthesis 1 minus the square root of 3 close parenthesis")[$− 2 ( 1 − sqrt(3) )$]


        
      
      
        #math.equation(block: true, alt: "the fraction 3 over 7 minus the square root of 2")[$display(frac(3, 7 − sqrt(2)))$]


      
      
        #math.equation(block: true, alt: "the fraction x over x minus the square root of 3")[$display(frac(x, x − sqrt(3)))$]


        
          #math.equation(block: true, alt: "the fraction x open parenthesis x plus the square root of 3 close parenthesis over x squared minus 3")[$display(frac(x ( x + sqrt(3) ), x^(2) − 3))$]


        
      
      
        #math.equation(block: true, alt: "the fraction y over the square root of 5 minus y")[$display(frac(y, sqrt(5) − y))$]


      
      
        #math.equation(block: true, alt: "the fraction the square root of 6 minus 3 over 2 minus the square root of 6")[$display(frac(sqrt(6) − 3, 2 − sqrt(6)))$]


        
          #math.equation(block: true, alt: "the fraction the square root of 6 over 2")[$display(frac(sqrt(6), 2))$]


        
      
      
        #math.equation(block: true, alt: "the fraction the square root of x plus the square root of y over the square root of x minus the square root of y")[$display(frac(sqrt(x) + sqrt(y), sqrt(x) − sqrt(y)))$]


      
      
        Use your calculator to graph each function, and explain the result.


        + #math.equation(block: false, alt: "y equals the square root of x squared")[$y = sqrt(x^(2))$]
+ #math.equation(block: false, alt: "y equals the cube root of x cubed")[$y = root(3, x^(3))$]


        
          + #math.equation(block: false, alt: "y equals the square root of x squared equals vertical bar x vertical bar")[$y = sqrt(x^(2)) = | x |$]
+ #math.equation(block: false, alt: "y equals the cube root of x cubed equals x")[$y = root(3, x^(3)) = x$]


        
      
      
        Use your calculator to graph each function, and explain the result.


        + #math.equation(block: false, alt: "y equals open parenthesis x to the power 4 close parenthesis to the power 1 / 4")[$y = ( x^(4) )^(1 / 4)$]
+ #math.equation(block: false, alt: "y equals open parenthesis x to the power 5 close parenthesis to the power 1 / 5")[$y = ( x^(5) )^(1 / 5)$]


      
      For Problems 53-54, do not assume that variables represent positive numbers. Use absolute value bars as necessary to simplify the radicals.


      
        + #math.equation(block: false, alt: "the square root of 4 x squared")[$sqrt(4 x^(2))$]
+ #math.equation(block: false, alt: "the square root of open parenthesis x minus 5 close parenthesis squared")[$sqrt(( x − 5 )^(2))$]
+ #math.equation(block: false, alt: "the square root of x squared minus 6 x plus 9")[$sqrt(x^(2) − 6 x + 9)$]


        
          + #math.equation(block: false, alt: "2 vertical bar x vertical bar")[$2 | x |$]
+ #math.equation(block: false, alt: "vertical bar x minus 5 vertical bar")[$| x − 5 |$]
+ #math.equation(block: false, alt: "vertical bar x minus 3 vertical bar")[$| x − 3 |$]


        
      
      
        + #math.equation(block: false, alt: "the square root of 9 x squared y to the power 4")[$sqrt(9 x^(2) y^(4))$]
+ #math.equation(block: false, alt: "the square root of open parenthesis 2 x minus 1 close parenthesis squared")[$sqrt(( 2 x − 1 )^(2))$]
+ #math.equation(block: false, alt: "the square root of 9 x squared minus 6 x plus 1")[$sqrt(9 x^(2) − 6 x + 1)$]


      
      For Problems 55-78, solve


      
        #math.equation(block: true, alt: "the square root of x minus 5 equals 3")[$sqrt(x) − 5 = 3$]


        
          #math.equation(block: true, alt: "64")[$64$]


        
      
      
        #math.equation(block: true, alt: "the square root of x minus 4 equals 1")[$sqrt(x) − 4 = 1$]


      
      
        #math.equation(block: true, alt: "the square root of y plus 6 equals 2")[$sqrt(y + 6) = 2$]


        
          #math.equation(block: true, alt: "minus 2")[$− 2$]


        
      
      
        #math.equation(block: true, alt: "the square root of y minus 3 equals 5")[$sqrt(y − 3) = 5$]


      
      
        #math.equation(block: true, alt: "4 the square root of z minus 8 equals minus 2")[$4 sqrt(z) − 8 = − 2$]


        
          #math.equation(block: true, alt: "the fraction 9 over 4")[$display(frac(9, 4))$]


        
      
      
        #math.equation(block: true, alt: "minus 3 the square root of z plus 14 equals 8")[$− 3 sqrt(z) + 14 = 8$]


      
      
        #math.equation(block: true, alt: "5 plus 2 the square root of 6 minus 2 w equals 13")[$5 + 2 sqrt(6 − 2 w) = 13$]


        
          #math.equation(block: true, alt: "minus 5")[$− 5$]


        
      
      
        #math.equation(block: true, alt: "8 minus 3 the square root of 9 plus 2 w equals minus 7")[$8 − 3 sqrt(9 + 2 w) = − 7$]


      
      
        #math.equation(block: true, alt: "3 z plus 4 equals the square root of 3 z plus 10")[$3 z + 4 = sqrt(3 z + 10)$]


        
          #math.equation(block: true, alt: "the fraction minus 1 over 3")[$display(frac(− 1, 3))$]


        
      
      
        #math.equation(block: true, alt: "2 x minus 3 equals the square root of 7 x minus 3")[$2 x − 3 = sqrt(7 x − 3)$]


      
      
        #math.equation(block: true, alt: "2 x plus 1 equals the square root of 10 x plus 5")[$2 x + 1 = sqrt(10 x + 5)$]


        
          #math.equation(block: true, alt: "the fraction minus 1 over 2 , 2")[$display(frac(− 1, 2)) , " " 2$]


        
      
      
        #math.equation(block: true, alt: "4 x plus 5 equals the square root of 3 x plus 4")[$4 x + 5 = sqrt(3 x + 4)$]


      
      
        #math.equation(block: true, alt: "the square root of y plus 4 equals y minus 8")[$sqrt(y + 4) = y − 8$]


        
          #math.equation(block: true, alt: "12")[$12$]


        
      
      
        #math.equation(block: true, alt: "4 the square root of x minus 4 equals x")[$4 sqrt(x − 4) = x$]


      
      
        #math.equation(block: true, alt: "the square root of 2 y minus 1 equals the square root of 3 y minus 6")[$sqrt(2 y − 1) = sqrt(3 y − 6)$]


        
          #math.equation(block: true, alt: "5")[$5$]


        
      
      
        #math.equation(block: true, alt: "the square root of 4 y plus 1 equals the square root of 6 y minus 3")[$sqrt(4 y + 1) = sqrt(6 y − 3)$]


      
      
        #math.equation(block: true, alt: "the square root of x minus 3 the square root of x equals 2")[$sqrt(x − 3) sqrt(x) = 2$]


        
          #math.equation(block: true, alt: "4")[$4$]


        
      
      
        #math.equation(block: true, alt: "the square root of x the square root of x minus 5 equals 6")[$sqrt(x) sqrt(x − 5) = 6$]


      
      
        #math.equation(block: true, alt: "the square root of y plus 4 equals the square root of y plus 20 minus 2")[$sqrt(y + 4) = sqrt(y + 20) − 2$]


        
          #math.equation(block: true, alt: "5")[$5$]


        
      
      
        #math.equation(block: true, alt: "4 the square root of y plus the square root of 1 plus 16 y equals 5")[$4 sqrt(y) + sqrt(1 + 16 y) = 5$]


      
      
        #math.equation(block: true, alt: "the square root of x plus the square root of 2 equals the square root of x plus 2")[$sqrt(x) + sqrt(2) = sqrt(x + 2)$]


        
          #math.equation(block: true, alt: "0")[$0$]


        
      
      
        #math.equation(block: true, alt: "the square root of 4 x plus 17 equals 4 minus the square root of x plus 1")[$sqrt(4 x + 17) = 4 − sqrt(x + 1)$]


      
      
        #math.equation(block: true, alt: "the square root of 5 plus x plus the square root of x equals 5")[$sqrt(5 + x) + sqrt(x) = 5$]


        
          #math.equation(block: true, alt: "4")[$4$]


        
      
      
        #math.equation(block: true, alt: "the square root of y plus 7 plus the square root of y plus 4 equals 3")[$sqrt(y + 7) + sqrt(y + 4) = 3$]


      
      
        Explain why the following first step for solving the radical equation is incorrect:


        #math.equation(block: true, alt: "the square root of x minus 5 plus the square root of 2 x minus 1, equals 8; open parenthesis x minus 5 close parenthesis plus open parenthesis 2 x minus 1 close parenthesis, equals 64")[$sqrt(x − 5) + sqrt(2 x − 1) & = 8 \ ( x − 5 ) + ( 2 x − 1 ) & = 64$]
        
          We cannot square each term separately; we must square each side of the equation.


        
      
      
        Explain why the following first step for solving the radical equation is incorrect:


        #math.equation(block: true, alt: "the square root of x plus 2 plus 1, equals the square root of 2 x minus 3; open parenthesis x plus 2 close parenthesis plus 1, equals 2 x minus 3")[$sqrt(x + 2) + 1 & = sqrt(2 x − 3) \ ( x + 2 ) + 1 & = 2 x − 3$]
      
      For Problems 81-84, write the complex fraction as a simple fraction in lowest terms, and rationalize the denominator.


      
        #math.equation(block: true, alt: "the fraction the fraction 2 over the square root of 7 over 1 minus the fraction the square root of 3 over the square root of 7")[$display(frac(display(frac(2, sqrt(7))), 1 − display(frac(sqrt(3), sqrt(7)))))$]


        
          #math.equation(block: true, alt: "the fraction the square root of 7 plus the square root of 3 over 2")[$display(frac(sqrt(7) + sqrt(3), 2))$]


        
      
      
        #math.equation(block: true, alt: "the fraction the fraction 1 over 4 over the fraction the square root of 5 over 2 the square root of 2 minus the fraction the square root of 3 over 2 the square root of 2")[$display(frac(display(frac(1, 4)), display(frac(sqrt(5), 2 sqrt(2))) − display(frac(sqrt(3), 2 sqrt(2)))))$]


      
      
        #math.equation(block: true, alt: "the fraction the fraction the square root of 3 over 2 plus the fraction 1 over the square root of 2 over 1 minus the fraction the square root of 3 over 2 times the fraction 1 over the square root of 2")[$display(frac(display(frac(sqrt(3), 2)) + display(frac(1, sqrt(2))), 1 − display(frac(sqrt(3), 2)) ⋅ display(frac(1, sqrt(2)))))$]


        
          #math.equation(block: true, alt: "the fraction 6 the square root of 3 plus 7 the square root of 2 over 5")[$display(frac(6 sqrt(3) + 7 sqrt(2), 5))$]


        
      
      
        #math.equation(block: true, alt: "the fraction the fraction 1 over the square root of 3 minus the fraction the square root of 5 over 3 over 1 plus the fraction 1 over the square root of 3 times the fraction the square root of 5 over 3")[$display(frac(display(frac(1, sqrt(3))) − display(frac(sqrt(5), 3)), 1 + display(frac(1, sqrt(3))) ⋅ display(frac(sqrt(5), 3))))$]