Question

solve for X. please jaldi thoda​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: solve \: for :value(s) \: of \: x \\ \\ given \: quadratic \: equation \: is \\ (a + b) {}^{2} x {}^{2} + 8(a {}^{2} - b {}^{2} )x + 16(a - b) {}^{2} = 0 \\ \\ comparing \: with \: ax {}^{2} + bx + c = 0 \\ we \: have \\ a = (a + b) {}^{2} \\ b = 8(a {}^{2} - b {}^{2} ) \\ c = 16(a - b) \\ \\ Δ = b {}^{2} - 4ac \\ =[ 8(a {}^{2} - b {}^{2} )] {}^{2} - 4 \times 16(a + b) {}^{2} (a - b) {}^{2} \\ = 64(a {}^{2} - b) {}^{2} - 64(a + b)(a - b)(a + b)(a - b) \\ = 64(a {}^{2} - b {}^{2} ) {}^{2} - 64(a {}^{2} - b {}^{2} )(a {}^{2} - b {}^{2} ) \\ = 64(a {}^{2} - b {}^{2} ) {}^{2} - 64(a {}^{2} - b {}^{2} ) {}^{2} \\ Δ= 0 \\ \\ we \: know \: that \\ by \: shreedharachrya \: method \\ \\ x = \frac{ - b \: ± \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ = \frac{ - 8(a {}^{2} - b {}^{2} )± \sqrt{0} }{2(a + b) {}^{2} } \\ \\ = \frac{ - 8( a + b)(a - b)}{2(a + b)(a + b)} \\ \\ = \frac{ - 4 [- (b - a)]}{a + b)} \\ \\ = \frac{4(b - a)}{a + b}$