Question

If α, β are the zeros of the polynomial f(x) = ax² + bx + c, then (1/α²) + (1/β²) =
A. (b² - 2ac)/a²
B. (b² - 2ac)/c²
C. (b² + 2ac)/a²
D. (b² + 2ac)/c²

Answer 1

Step-by-step explanation:

option C is correct answer

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: ANSWER \: \: } \mid}}}}}$

$\: \: \: \: \: \: \: \tt{If \: \: \alpha \: \: and \: \: \beta \: \: are \: \: the \: \: zeros \: \: of } \\ \tt{the \: \: polynomial \: \: f(x) = ax{}{2} + bx + c ,} \\ \tt{then \: \frac{1}{\alpha {}^{2} } + \frac{1}{\beta {}^{2} } = } \\ \\ \sf{ A. \: \frac{b {}^{2} - 2ac}{a {}^{2} }} \\ \\ \sf{B. \: \frac{b {}^{2} - 2ac}{c {}^{2} } \: \: \: \: \: \: \huge{ \underline{ \: \: \checkmark \: \: }}} \\ \\ \sf{ C. \: \frac{b {}^{2} + 2ac}{a {}^{2} }} \\ \\ \sf{D.\frac{b {}^{2} + 2ac}{c {}^{2} }}</p><p>$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline {\mathfrak{ \: \: Given, \: \: }} \\ \\ \text{Quadratic \: \: polynomial \: \: is : } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ \red{f(x) = ax {}^{2} + bx + c.}}} \\ \\ \text{The zeroes of thr polynomial are : } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge{\tt{ \red{ \alpha \: \: and \: \: \beta }}} \\ \\ \underline {\mathfrak{ \: \:To \: \: find :- }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \huge{\tt{ \pink{ \frac{1}{ \alpha {}^{2} } + \frac{1}{ \beta {}^{2} } = \: ?}}}$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \: \ \\ \: \: \: \: \: \: \: \sf{ Sum \: \: of \: \: zeroes = \frac{ - b}{a} } \\ \\ \sf{ \implies \: \alpha + \beta =\frac{ - b}{a} \: \: - - - - - - > (1)} \\ \\ \underline{ \mathfrak{ \: And \: , }} \\ \\ \: \: \: \: \: \: \: \sf{Product \: \: of \: \: zeroes = \frac{c}{a}} \\ \\ \sf {\implies \: \alpha. \beta = \frac{c}{a}\: \: - - - - - - > (2)}$

$\mathfrak{Now,} \\ \\ \underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{a{}^2 + b{}^2 = (a + b){}^2 - 2ab}}$

$\tt{ \alpha {}^{2} + \beta {}^{2} = ( \alpha + \beta) {}^{2} - 2 .\alpha. \beta } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{= ( \frac{-b}{a}){}^2 - 2( \frac{c}{a}) \: \: \: \: \: \: \: \: \: \: \:[From (1) and (2)]} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{= \frac{b{}^2}{a{}^2}- \frac{2c}{a}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{= \frac{b{}^2 - 2ca}{a{}^2} \: \: \: \: \: --- - - - - - > (3)}$

$\: \: \: \: \: \: \: \: \sf{ \pink{\frac{1}{ \alpha{}^2} + \frac{1}{ \beta{}^2}}} \\ \\ \sf{= \frac{ \beta{}^2 + \alpha{}^2}{ \alpha {}^{2} . \beta{}^2} }\\ \\ \sf{= \frac{ \alpha {}^{2} + \beta {}^{2} }{ (\alpha. \beta ){}^{2} } } \\ \\ \sf{ = \frac{ \frac{b^2 - 2ca}{a^2 }}{( \frac{c}{a} ) {}^{2} } } \\ \\ \sf{ = \frac{ \frac{b^2 - 2ca}{a^2 }}{\frac{c {}^{2} }{a {}^{2} } } } \\ \\ \sf{= \frac{b^2 - 2ca}{ \cancel{a^2 }} \times \frac{ \cancel{a {}^{2} }}{c {}^{2} }} \\ \\ \sf{ = \frac{ (b{}^2 - 2ca)}{c{}^2}} \\ \\ \sf{ = \pink{ \frac{ (b{}^2 - 2ac)}{c{}^2}}}$