Question

if alpha and beta are the zeros of the quadratic polynomial f(x) = x^2-4x+3 . find the value of alpha^4*beta^2+alpha^2*beta^4

Answer 1

it's basic question of quadratic equation.

I hope you got it

Answer 2

Answer: The value of $\alpha^4 \beta^2+\alpha^2\beta^4$ is 90

Step-by-step explanation:                  

Given: $f(x)= x^2 - 4x+3$

To find : $\alpha^4 \beta^2+\alpha^2\beta^4$

Now comparing the standard polynomial $f(x)= ax^2 + bx+c$

we have a= 1, b= -4 , c= 3

Sum of zeroes= $\alpha + \beta = \dfrac{-b}{a} =\dfrac{-(-4)}{1}=4$

Product of zeroes = $\alpha \beta = \dfrac{c}{a} = \dfrac{3}{1} =3$

now

$\alpha^4 \beta^2+\alpha^2\beta^4= \alpha^2\beta^2(\alpha^2+ \beta^2)\\\\= (\alpha\beta)^2((\alpha+ \beta)^2-2\alpha\beta)\\\\= 3^2( 4^2-2\times 3)\\\\=9(16-6)\\\\=9\times 10=90$

Hence, the value of $\alpha^4 \beta^2+\alpha^2\beta^4$ is 90