Question

if alpha and beta are the zeros of the quadratic polynomial f(x)= ax²+bx+c, then evaluate alpha⁴ +beta⁴

Answer 1

$Given \: \alpha \:and \:beta \:are \: zeroes \\of \: the \: Quadratic \: polynomial \:f(x) = ax^{2}+bx+c$

$i) Sum \:of \:the \: zeroes = \frac{-b}{a}$

$\implies \alpha + \beta = \frac{-b}{a}\: --(1)$

$ii) Product \:of \:the \: zeroes = \frac{c}{a}$

$\implies \alpha \beta = \frac{c}{a}\: --(2)$

$Now, iii ) \alpha^{2} + \beta^{2} \\= \big(\alpha + \beta \big)^{2} - 2\alpha \beta \\= \Big(\frac{-b}{a} \Big)^{2} - 2\times \frac{c}{a} \\= \frac{b^{2}}{a^{2}} - \frac{2c}{a} \\= \frac{b^{2}-2ac}{a^{2}} \: --(3)$

$\red { Value \:of \: \alpha^{4} + \beta^{4}} \\= \big(\alpha^{2} + \beta^{2} \big)^{2} - 2\alpha^{2} \beta^{2} \\=\Big( \frac{b^{2}-2ac}{a^{2}}\Big)^{2} - 2 \times (\alpha \beta)^{2}\\= \frac{\Big(b^{2}-2ac\Big)^{2}}{a^{4}} - 2\times \Big(\frac{c}{a}\Big)^{2} \\= \frac{\Big(b^{2}-2ac\Big)^{2}}{a^{4}} - 2\times \frac{c^{2}}{a^{2}} \\= \frac{(b^{2}-2ac)^{2} - 2a^{2}c^{2}}{a^{4}}$

Therefore.,

$\red { Value \:of \: \alpha^{4} + \beta^{4}}\\\green { = \frac{(b^{2}-2ac)^{2} - 2a^{2}c^{2}}{a^{4}}}$

•••♪

Answer 2

a⁴+b⁴

= (a²+b²)²-2a²b²

= ((a+b)²-2ab)²-2a²b²

= ((- b/a)²-2c/a)² - 2(c/a)²

= (b²/a²-2c/a)²-2c²/a²

= (b²-2ac)²/a⁴-2c²/a²

= ((b²-2ac)²-2a²c²)/a⁴

= (b⁴+4a²c²-4b²ac - 2a²c²)/a⁴

Answer

= (b⁴+2a²c²-4ab²)/a⁴

BY VEDHANT KHAJURIA