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Zorluk: OrtaSurds and Indices

If x=7+57−5x = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} and y=7−57+5y = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}, what is the value of x2+y2+xyx^2 + y^2 + xy?

  1. A
    142142
  2. 143143Cevap
  3. C
    144144
  4. D
    145145

Cevap

The value of x2+y2+xyx^2 + y^2 + xy is 143143.
Rationalizing both surds gives x=6+35x = 6 + \sqrt{35} and y=6−35y = 6 - \sqrt{35}. The sum x+y=12x + y = 12 and product xy=1xy = 1. Substituting into the algebraic identity x2+y2+xy=(x+y)2−xyx^2 + y^2 + xy = (x + y)^2 - xy gives 122−1=14312^2 - 1 = 143.

Adım Adım Çözüm

1
Rationalize the denominator for xx and yy
x=(7+5)2(7)2−(5)2=7+5+2352=6+35x = \frac{(\sqrt{7} + \sqrt{5})^2}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{7 + 5 + 2\sqrt{35}}{2} = 6 + \sqrt{35}, and similarly y=6−35y = 6 - \sqrt{35}.
Eliminating radicals from denominators simplifies calculations.
2
Compute the sum (x+y)(x + y) and product xyxy
x+y=(6+35)+(6−35)=12x + y = (6 + \sqrt{35}) + (6 - \sqrt{35}) = 12, and xy=(6+35)(6−35)=36−35=1xy = (6 + \sqrt{35})(6 - \sqrt{35}) = 36 - 35 = 1.
Using symmetric expressions simplifies evaluating degree 2 polynomials.
3
Express x2+y2+xyx^2 + y^2 + xy in terms of (x+y)(x + y) and xyxy
x2+y2+xy=(x+y)2−xy=122−1=144−1=143x^2 + y^2 + xy = (x + y)^2 - xy = 12^2 - 1 = 144 - 1 = 143.
Applying the identity x2+y2=(x+y)2−2xyx^2 + y^2 = (x + y)^2 - 2xy gives x2+y2+xy=(x+y)2−xyx^2 + y^2 + xy = (x + y)^2 - xy.

Anahtar Kavram

Rationalization of surds and application of algebraic identities
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Konular

Surds and Indices

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