Question

solve for x- (a+b)x²-4abx-(a-b)²=0​

Answer 1

your solution is as follows

pls mark it as brainliest

$to \: solve \: for : \\ value(s) \: of \: x \\ \\ (a + b) {}^{2} x {}^{2} - 4abx - (a - b) {}^{2} = 0 \\ comparing \: above \\ \: quadratic \: equation \: with \\ ax {}^{2} + bx + c = 0 \\ we \: get \\ \\ a = (a + b) {}^{2} \\ b = - 4ab \\ c = - (a - b) {}^{2}$

$we \: know \: that \\ Δ= b {}^{2} - 4ac \\ \\ = (4ab) {}^{2} - 4(a + b) {}^{2} ( - (a - b) {}^{2} ) \\ \\ = 16a {}^{2} b {}^{2} + 4(a + b) {}^{2} (a - b) {}^{2} \\ \\ = 4[4a {}^{2} b {}^{2} + (a + b)(a + b)(a - b)(a - b)] \\ \\ = 4[4a {}^{2} b {}^{2} + (a {}^{2} - b {}^{2} ) {}^{2} ] \\ \\ = 4(4a {}^{2} b {}^{2} + a {}^{4} + b {}^{4} - 2a {}^{2} b {}^{2} \\ \\ = 4(a {}^{4} + b {}^{4} + 2a {}^{2} b {}^{2} ) \\ \\ Δ = 4(a {}^{2} + b {}^{2} ) {}^{2}$

$so \: by \: using \: formula \: method \\ we \: get \\ \\ x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - ( - 4ab)± \sqrt{4(a {}^{2} + b {}^{2} ) {}^{2} } }{2(a + b) {}^{2} } \\ \\ = \frac{4ab±2(a {}^{2} + b {}^{2}) }{2(a + b) {}^{2} } \\ \\ = \frac{2(2ab + a {}^{2} + b {}^{2}) }{2(a + b) {}^{2} } \: or \: = \frac{ - 2( a {}^{2} + b {}^{2} - 2ab)}{2(a + b) {}^{2} } \\ \\ = \frac{(a + b) {}^{2} }{(a + b) {}^{2} } \: \: or \: \: = - \frac{(a - b) {}^{2} }{(a + b) {}^{2} } \\ \\ x = 1 \: \: or \: \: x = ( \frac{a - b}{a + b} ) {}^{2}$