If p and q are the roots of the equation ax^2 + bx + c = 0, find the value of 1/(ap^2+c)^2 + 1/(aq^2+c)^2.

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Solution

$I f p a n d q a r e r o o t s o f t h e e q u a t i o n a x^{2} + b x + c = 0 t h e n a p^{2} + b p + c = 0 a p^{2} + c = - b p a q^{2} + b q + c = 0 a q^{2} + c = - b q \frac{1}{{(a p^{2} + c)}^{2}} + \frac{1}{{(a q^{2} + c)}^{2}} = \frac{1}{{(- b p)}^{2}} + \frac{1}{{(- b q)}^{2}} = \frac{1}{b^{2}} \{\frac{1}{p^{2}} + \frac{1}{q^{2}}\} = \frac{1}{b^{2}} \{\frac{p^{2} + q^{2}}{p^{2} q^{2}}\} = \frac{1}{b^{2}} \{\frac{p^{2} + q^{2}}{{(p q)}^{2}}\} = \frac{1}{b^{2}} \{\frac{(p^{2} + q^{2} + 2 p q) - 2 p q}{{(p q)}^{2}}\} = \frac{1}{b^{2}} \{\frac{{(p + q)}^{2} - 2 p q}{{(p q)}^{2}}\} U \sin g f o r m u l a f o r s u m a n d p r o d u c t o f r o o t s p + q = - \frac{b}{a} p q = \frac{c}{a} = \frac{1}{b^{2}} \{\frac{{(- \frac{b}{a})}^{2} - 2 \frac{c}{a}}{{(\frac{c}{a})}^{2}}\} = \frac{1}{b^{2}} \{\frac{\frac{b^{2}}{a^{2}} - \frac{2 c}{a}}{\frac{c^{2}}{a^{2}}}\} = \frac{1}{b^{2}} \{\frac{\frac{b^{2} - 2 a c}{a^{2}}}{\frac{c^{2}}{a^{2}}}\} = \frac{1}{b^{2}} \{\frac{b^{2} - 2 a c}{c^{2}}\} = \frac{b^{2} - 2 a c}{b^{2} c^{2}}$

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