Question

If alpha and beta are zeros of the quadratic polynomial 4x^2-5x+1 then find:
i)
$\alpha {}^{4} + \beta {}^{4}$
ii)
$\frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta$
​

Answer 1

Given

if $\alpha\:&\beta$ are zeros of the quadratic polynomial 4x²-5x+1

To find

1.$\alpha {}^{4} + \beta {}^{4}$

2.$\frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta$

Solution

1. $\alpha + \beta=\frac{-b}{a}$

$\alpha + \beta=\frac{5}{4}$

&

$\implies\alpha\times \beta=\frac{c}{a}$

$\implies\alpha\times \beta=\frac{1}{4}$

Now,

$\alpha {}^{4} + \beta {}^{4}$

$\implies\alpha {}^{4} + \beta {}^{4}=(\alpha^2 + \beta^2)^2-2\alpha^2\times \beta^2$

Putting the values

$\implies \alpha {}^{4} + \beta {}^{4}$=$\frac{5}{4}-2 (\alpha\times \beta)^2$

$\implies \alpha {}^{4} + \beta {}^{4}$=$\frac{5}{4}-2 (\frac{1}{4})^2$

$\implies \alpha {}^{4} + \beta {}^{4}$=$\frac{5}{4}- (\frac{1}{8})$

$\implies \alpha {}^{4} + \beta {}^{4}$=$\frac{10-1}{8}$

$\implies \alpha {}^{4} + \beta {}^{4}$=$\frac{9}{8}$

2.

$\frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta$

$\implies \frac{\alpha+\beta-\alpha^2\beta^2 }$

$\implies \frac{\alpha+\beta-\alpha^2\beta^2 }$=$\frac{20-1}{4}$

$\implies \frac{\alpha+\beta-\alpha^2\beta^2 }$=$\frac{19}{4}$