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Difficulty: HardExponents, Radicals, and Algebraic Expressions

If x=7−43x = \sqrt{7 - 4\sqrt{3}} and y=7+43y = \sqrt{7 + 4\sqrt{3}}, what is the value of x−2+y−2(x+y)−1\frac{x^{-2} + y^{-2}}{(x+y)^{-1}}?

  1. A
    14\frac{1}{4}
  2. B
    72\frac{7}{2}
  3. C
    1414
  4. 5656Answer
  5. E
    6464

Answer

5656
The expression x=7−43x = \sqrt{7 - 4\sqrt{3}} can be unnested by recognizing 7−437 - 4\sqrt{3} as the perfect square (2−3)2(2 - \sqrt{3})^2, so x=2−3x = 2 - \sqrt{3} and y=2+3y = 2 + \sqrt{3}. This gives x+y=4x+y = 4 and xy=1xy = 1. The numerator x−2+y−2=x2+y2(xy)2=(x+y)2−2xy1=14x^{-2} + y^{-2} = \frac{x^2+y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{1} = 14. The denominator (x+y)−1=14(x+y)^{-1} = \frac{1}{4}. Dividing 1414 by 14\frac{1}{4} yields 5656.

Step-by-Step Solution

1
Simplify the radical expressions for xx and yy.
x=2−3x = 2 - \sqrt{3} and y=2+3y = 2 + \sqrt{3}.
Note that 7−43=4−43+3=(2−3)27 - 4\sqrt{3} = 4 - 4\sqrt{3} + 3 = (2 - \sqrt{3})^2, so 7−43=2−3\sqrt{7 - 4\sqrt{3}} = 2 - \sqrt{3}. Similarly, 7+43=(2+3)27 + 4\sqrt{3} = (2 + \sqrt{3})^2.
2
Find the sum x+yx+y and product xyxy.
x+y=4x+y = 4 and xy=1xy = 1.
(2−3)+(2+3)=4(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4, and (2−3)(2+3)=22−(3)2=4−3=1(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1.
3
Simplify the numerator x−2+y−2x^{-2} + y^{-2}.
x−2+y−2=14x^{-2} + y^{-2} = 14.
x−2+y−2=1x2+1y2=x2+y2(xy)2=(x+y)2−2xy(xy)2=42−2(1)12=14x^{-2} + y^{-2} = \frac{1}{x^2} + \frac{1}{y^2} = \frac{x^2 + y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{(xy)^2} = \frac{4^2 - 2(1)}{1^2} = 14.
4
Evaluate the full expression x−2+y−2(x+y)−1\frac{x^{-2} + y^{-2}}{(x+y)^{-1}}.
141/4=56\frac{14}{1/4} = 56.
The denominator is (x+y)−1=(4)−1=14(x+y)^{-1} = (4)^{-1} = \frac{1}{4}. Dividing 1414 by 14\frac{1}{4} equals 14⋅4=5614 \cdot 4 = 56.

Key Concept

Nested Radical Simplification & Algebraic Exponent Identities

Alternative Method

Instead of unnesting the radicals first, observe that x2=7−43x^2 = 7 - 4\sqrt{3} and y2=7+43y^2 = 7 + 4\sqrt{3}. Then x2y2=(7−43)(7+43)=49−48=1x^2 y^2 = (7-4\sqrt{3})(7+4\sqrt{3}) = 49 - 48 = 1, and x2+y2=14x^2 + y^2 = 14. Thus x−2+y−2=x2+y2x2y2=14x^{-2} + y^{-2} = \frac{x^2+y^2}{x^2 y^2} = 14. Next, find (x+y)2=x2+y2+2xy=14+2(1)=16(x+y)^2 = x^2 + y^2 + 2xy = 14 + 2(1) = 16, so x+y=4x+y = 4. Then (x+y)−1=14(x+y)^{-1} = \frac{1}{4}, leading directly to 141/4=56\frac{14}{1/4} = 56.
Estimated Time:2m 0s
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Topics

Exponents, Radicals, and Algebraic Expressions

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