Question

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Answer 1

Answer:

3rd question Answer I think this helpful for you

Answer 2

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: here \: given \: quadratic \: equation \: is \: \\ (p - q) {}^{2} .x {}^{2} \: + 2(p { }^{2} - q {}^{2} ).x + k = 0 \\ on \: comparing \: this \: quadratic \: equation \: with \: ax {}^{2} + bx + c = 0 \\ we \: get \\ a = (p - q) {}^{2} \\ b = 2(p {}^{2} - q { }^{2} ) \\ c = k$

$now \: let \: its \: roots \: be \: \alpha \: and \: \beta \: respectively \\ so \: here \: \alpha = \beta \: \: \: \: \: \: \: (given) \\ \\ so \: we \: know \: that \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha + \alpha = \frac{ - 2(p {}^{2} - q {}^{2} )}{(p - q) {}^{2} } \\ \\ 2 \alpha = \frac{ - 2(p + q)(p - q)}{(p - q)(p - q)} \\ \\ \alpha = \frac{ - (p + q)}{(p - q)} \: \: \: \: \: \: (1)$

$similarly \: we \: know \: that \\ \alpha \beta = \frac{c}{a} \\ \\ \alpha . \alpha = \frac{k}{(p - q) {}^{2} } \\ \\ \alpha {}^{2} = \frac{k}{(p - q) {}^{2} } \\ \\ ( \frac{ - (p + q)}{p - q)} ) {}^{2} = \frac{k}{(p - q) {}^{2} } \\ \\ ( \frac{p + q}{p - q} ) {}^{2} = \frac{k}{(p - q) {}^{2} } \\ \\ \frac{(p + q) {}^{2} }{(p - q) {}^{2} } = \frac{k}{(p - q) {}^{2} } \\ \\ k = (p + q) {}^{2}$