Question

If the sum of the root of the quadratic equations is 1/x+p+1/x+q=1/r is 0 show that the product of root is -p^p+q^q/2

Answer 1

Answer:

$\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ $\\ \frac{x + q + x + p}{(x + p)(x + q} = \frac{1}{r}$ $\\ \frac{2x + p + q}{ {x }^{2} + qx + px + pq} = \frac{1}{r}$ $\\ r(2x + p + q) = {x}^{2} + qx + px + pq$ $\\ 2rx + pr + qr = {x}^{2} + qx + px + pq$ $\\ 0 = {x}^{2} + qx + px + pq - 2rx + - pr - qr$ $\\ {x}^{2} + qx + px - 2rx + pq - pr - qr = 0$ $\\ {x}^{2} + x(q + p - 2r) + pq - pr - qr = 0$ $\\ {x}^{2} + (p + q - 2r)x + (pq - pr - qr) = 0........(1)$ $\\ Comparing \: with \: ax ^{2} + bx + c = 0 \: we \: get$ $\\ a = 1 \\ b = p + q - 2r \\ c = pq - pr - qr$ $\\ let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: equation \: 1$ $\\ let \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ -(p + q - 2r)}{1} \\ \\ \alpha + \beta = - p - q + 2r$ $\\ but \: \alpha + \beta = 0.......(given)$ $\\ - p + q - 2r = 0 \\ \\ 2r = p + q$ $\\ \\ r = \frac{p + q}{2} \\ \\ also \: \alpha \beta = \frac{c}{a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{pq - pr - qr}{1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = pq - pr - qr \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = pq - r(p + q)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = pq - ( \frac{p + q}{2})(p + q).........r = \frac{p + q}{2} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = pq - \frac{ {(p + q)}^{2} }{2}$ $\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2pq - (p + q )^{2} }{2} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2pq - (p ^{2} + 2pq + {q}^{2}) }{2} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2pq - {p}^{2} - 2pq - {q}^{2} }{2} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - {p}^{2} - {q}^{2} }{2}$ $\\ \\ \alpha \beta = - ( \frac{ {p}^{2} + {q}^{2} }{2} )$

Answer 2

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