Question

Difficulty: MediumExponents, Radicals, and Algebraic Expressions

If x=3+2x = \sqrt{3} + \sqrt{2}, what is the value of x4+x4x^4 + x^{-4}?

  1. 98Answer
  2. B
    100
  3. C
    102
  4. D
    144
  5. E
    14

Answer

98
Rationalizing the denominator gives x1=32x^{-1} = \sqrt{3} - \sqrt{2}. Summing x+x1x + x^{-1} yields 232\sqrt{3}. Squaring this sum gives x2+2+x2=12x^2 + 2 + x^{-2} = 12, so x2+x2=10x^2 + x^{-2} = 10. Squaring once more gives x4+2+x4=100x^4 + 2 + x^{-4} = 100, which simplifies to x4+x4=98x^4 + x^{-4} = 98.

Step-by-Step Solution

1
Find the reciprocal expression x1x^{-1}
x1=13+2=32(3+2)(32)=32x^{-1} = \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \sqrt{3} - \sqrt{2}
Rationalizing the denominator simplifies the expression for x1x^{-1}.
2
Calculate the sum x+x1x + x^{-1}
x+x1=(3+2)+(32)=23x + x^{-1} = (\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3}
Adding the original expression and its reciprocal eliminates the 2\sqrt{2} term.
3
Square x+x1x + x^{-1} to find x2+x2x^2 + x^{-2}
(x+x1)2=x2+2(x)(x1)+x2=x2+2+x2=(23)2=12    x2+x2=10(x + x^{-1})^2 = x^2 + 2(x)(x^{-1}) + x^{-2} = x^2 + 2 + x^{-2} = (2\sqrt{3})^2 = 12 \implies x^2 + x^{-2} = 10
Using the algebraic identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 allows finding x2+x2x^2 + x^{-2}.
4
Square x2+x2x^2 + x^{-2} to find x4+x4x^4 + x^{-4}
(x2+x2)2=x4+2(x2)(x2)+x4=x4+2+x4=102=100    x4+x4=98(x^2 + x^{-2})^2 = x^4 + 2(x^2)(x^{-2}) + x^{-4} = x^4 + 2 + x^{-4} = 10^2 = 100 \implies x^4 + x^{-4} = 98
Squaring the second-power sum and subtracting the cross-term 2 yields the required value.

Key Concept

Rationalizing radical denominators and binomial squaring of reciprocals
Estimated Time:1m 30s
Rate this question