(solved)Question 22 SSC-CGL 2020 March 3 Shift 1

If x⁴ + x²y² + y⁴ = 21, x² + xy + y² = 7, then the value of (1/x² + 1/y²) = ?
(Rev. 31-Oct-2024)

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Question 22
SSC-CGL 2020 Mar 3 Shift 1

If x⁴ + x²y² + y⁴ = 21, x² + xy + y² = 7, then the value of (1/x² + 1/y²) = ?

Solution in Brief

Factorize x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² - xy + y²), to get x² - xy + y² = 3. Subtract it from x² + xy + y² = 7 to get xy = 2. Make a guess x = 1, y = 2 to get 5/4 as the answer!

Solution in Detail

Given x2+xy+y2=7\displaystyle x^2 + xy + y^2 = 7

(x2+y2)=7xy . . . (1)\displaystyle \therefore (x^2 + y^2) = 7 - xy\text{ . . . (1)}

Also given x4+x2y2+y4=21\displaystyle x^4 + x^2y^2 + y^4 = 21

Add x2y2\displaystyle x^2y^2 to both sides:

x4+2x2y2+y4=21+x2y2\displaystyle x^4 + 2x^2y^2 + y^4 = 21 + x^2y^2

Completing the squares

(x2+y2)2=21+x2y2\displaystyle (x^2 + y^2)^2 = 21 + x^2y^2

Putting x2+y2\displaystyle x^2 + y^2 from (1),

(7xy)2=21+x2y2\displaystyle (7 - xy)^2 = 21 + x^2y^2

    xy=2\displaystyle \implies xy = 2

Put in (1), x2+y2=72=5\displaystyle x^2 + y^2 = 7 - 2 = 5

To find 1x2+1y2\displaystyle \frac{1}{x^2} + \frac{1}{y^2}

=x2+y2(xy)2=522=54Ans\displaystyle = \frac{x^2 + y^2}{(xy)^2} = \frac{5}{2^2} = \frac54\:\underline{Ans}

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