Question

If p and q are the roots of the equation ax²+bx+c = 0 then find the value of 1/(ap²+c)² + 1/(aq²+c)²​

Answer 1

GivEn:

• p and q are the roots of the equation ax²+bx+c = 0

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To find:

• Value of $\sf \dfrac{1}{(ap^2 + c)^2} + \dfrac{1}{(aq^2 + c)^2}$

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SoluTion:

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As we know that,

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${\underline{\sf{\bigstar\;Sum\:of\:roots\;:}}}$

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$\sf p + q = \dfrac{- b}{a}$

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${\underline{\sf{\bigstar\;Product\:of\:roots\;:}}}$

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$\sf pq = \dfrac{c}{a}$

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Therefore,

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If p in the root of equation ax² + bx + c = 0 then,

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$:\implies\sf ap^2 + bp + c = 0$

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$:\implies\sf ap^2 + c = - bp$

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and

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If q in the root of equation ax² + bx + c = 0 then,

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$:\implies\sf aq^2 + bq + c = 0$

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$:\implies\sf aq^2 + c = - bq$

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Therefore,

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Value of $\sf \dfrac{1}{(ap^2 + c)^2} + \dfrac{1}{(aq^2 + c)^2}$ is,

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$:\implies\sf \dfrac{1}{( - bp)^2} + \dfrac{1}{( - bq)^2}$

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$:\implies\sf \dfrac{1}{b^2} \bigg( \dfrac{1}{p^2} + \dfrac{1}{q^2} \bigg)$

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$:\implies\sf \dfrac{1}{b^2} \bigg( \dfrac{p^2 + q^2}{(pq)^2} \bigg)$

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$:\implies\sf \dfrac{1}{b^2} \bigg( \dfrac{(p + q)^2 - 2pq}{(pq)^2} \bigg)$

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$:\implies\sf \dfrac{1}{b^2} \bigg( \dfrac{ \bigg( \frac{-b}{a} \bigg)^2 - 2 \bigg( \dfrac{c}{a} \bigg)}{ \bigg( \dfrac{c}{a} \bigg)^2} \bigg)$

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$:\implies\sf \dfrac{b^2 - 2ca}{(bc)^2}$

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${\underline{\underline{\bf{\pink{Hence,\;Solved!!}}}}}$

Answer 2

Answer:

b^2 - 2ac/b^2c^2

Step-by-step explanation:

explanation is given in the pic attached

@amirsalim95