Question

find quadratic equation such that its root are square of sum of the roots and square of difference of the roots of equation 2x^2+ 2 (p + q)x + p^2 +q^2 =0​

Answer 1

Answer: x² - 4pq x - ( p² - q²)² = 0

Step-by-step explanation:

2x² + 2(p+ q)x + p² + q² …. given equation with roots $\alpha$ & $\beta$

( $\alpha$ + β ) = {-2(p+ q)}/2 = - (p+ q)

=> ( $\alpha$ + β )² = (p+ q)² ……………. (1)

$\alpha$β = (p² + q² )/2

& ( $\alpha$ - β )² = ( $\alpha$ + β )² - 4£β

=> ( $\alpha$ - β )² = ( p+q)² - 4 * (p² + q² )/2

=> ( $\alpha$ - β )² = p² + q² + 2pq - 2p² - 2q² = - p² - q² + 2pq

=> ($\alpha$-β)² = - ( p- q)² ………….. (2)

Since, (1) & (2) are the roots of needed quadratic equation

=> equation is, ( x - (1)) ( x- (2)) = 0

So, {x- (p+ q)² } { x + ( p- q)² } = 0 is the required equation

=> x² - ( p+ q)² x + (p- q)²x - (p+ q)² ( p- q)² = 0

=> x² - ( p² + q² + 2pq - p² - q² + 2pq)x - ( p² - q² )² = 0

=> x² - 4pq x - ( p² - q²)² = 0

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