Question

ifα and β are zeros of polynomial f(x) = ax²+bx+c then find 1/α²+1/β²

Answer 1

Given Quadratic polynomial is ax^2 + bx + c.

Given a,b are the zeroes of the polynomial.

= > We know that sum of zeroes = -b/a

a + b = -b/a

= > We know that product of zeroes = c/a

 ab = c/a.


Now,

We know that a^2 + b^2 = (a + b)^2 - 2ab

= > (-b/a)^2 - 2(c/a)

= > (b^2/a^2) - 2c/a

= > (b^2 - 2ca)/a^2   ------- (1)


Given:

= > 1/a^2 + 1/b^2

= > a^2 + b^2/(ab)^2

= > (b^2 - 2ca)/a^2 * (a^2)/c^2

= > (b^2 - 2ca)/c^2



Hope this helps!

Answer 2

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

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

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

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

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

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