Question

qudratic equation sove​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest

Step-by-step explanation:

$to \: solve \: for = \\ x \: (in \: terms \: of \: a \: and \: b) \\ \\ so \: here \: given \: quadratic \: equation \: \: is \\ \\ \frac{b}{x - a} + \frac{a}{x - b} = 2 \\ \\ \frac{b(x - b) + a(x - a)}{(x - a)(x - b)} = 2 \\ \\ \\ bx - b {}^{2} + ax - a {}^{2} = 2(x {}^{2} - bx - ax + a b) \\ \\ = bx + ax - b {}^{2} - a {}^{2} = 2x {}^{2} - 2bx - 2ax + 2ab \\ \\ bx + 2bx + ax + 2ax = 2x {}^{2} + a {}^{2} + b {}^{2} + 2ab \\ \\ 2x {}^{2} + (a + b) {}^{2} = 3bx + 3ax \\ \\ 2x {}^{2} - 3 a x - 3bx + (a + b) {}^{2} = 0 \\ \\ = 2x {}^{2} - 3(a + b)x + (a + b) {}^{2} = 0$

$comparing \: this \: quadratic \: equation \\ with \: \: ax {}^{2} + bx + c = 0 \\ we \: have \\ \\ a = 2 \\ \\ b = - 3(a + b) \\ \\ c = (a + b) {}^{2}$

$so \: applying \: now \\ \\ Δ = b {}^{2} - 4ac \\ \\ = ( - 3(a + b)) {}^{2} - 4 \times 2 \times (a + b) {}^{2} \\ \\ = 9(a + b) {}^{2} - 8(a + b) {}^{2} \\ \\ = (a + b) {}^{2} (9 - 8) \\ \\ = (a + b) {}^{2} (1) \\ \\ Δ=( a + b) {}^{2}$

$so \: now \: by \: formula \: method \\ we \: get \\ \\ x = \frac{ - b \: ± \: \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - ( - 3(a + b) \: ± \: \sqrt{(a + b) {}^{2} } }{(2 \times 2)} \\ \\ \\ = \frac{3(a + b) \:± \: (a + b) }{4} \\ \\ = \frac{(a + b)(3 \:± \: 1) }{4} \\ \\ \\ x = \frac{(a + b)(3 + 1)}{4} \: \: or \: \: x = \frac{(a + b)(3 - 1)}{4} \\ \\ \\ = \frac{(a + b)(4)}{4} \: \: or \: \: = \frac{(a + b)(2)}{4} \\ \\ \\ x = a + b \: \: \: or \: \: \: x = \frac{a + b}{2}$