Question

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

Answer 1

Step-by-step explanation:

refer to the attachment ...☺☺☺☺☺

Answer 2

Answer:

α⁴ + β⁴ =  $\frac{b^4+2a^2c^2 - 4ab^2c}{a^4}$

Step-by-step explanation:

Given,

α and β are the zeros of the quadratic polynomial f(x) = ax²+bx+c

To find,

The value of α⁴ + β⁴

Solution:

Recall the concepts:

If α and β are the zeros of the  quadratic polynomial ax²+bx+c,

then Sum of zeros = α+β = $\frac{-b}{a}$

Product of zeros = αβ = $\frac{c}{a}$

α+β = $\frac{-b}{a}$ and  αβ = $\frac{c}{a}$ -----------------------(1)

α⁴ + β⁴ = (α² + β²)² - 2α²β²  ---------------(2)

α² + β² =(α+β)² - 2αβ

(α² + β²)² =((α+β)² - 2αβ)²

= (α+β)⁴ + 4α²β² - 2(α+β)² ×2αβ

= (α+β)⁴ + 4(αβ)² - 4(α+β)² ×αβ

(α² + β²)² =  (α+β)⁴ + 4(αβ)² - 4(α+β)² ×αβ

Equation (2) becomes, α⁴ + β⁴ =  (α+β)⁴ + 4(αβ)² - 4(α+β)² ×αβ - 2α²β²

= (α+β)⁴ + 4(αβ)² - 4(α+β)² ×αβ - 2(αβ)²

= (α+β)⁴ + 2(αβ)² - 4(α+β)² ×αβ

α⁴ + β⁴ = (α+β)⁴ + 2(αβ)² - 4(α+β)² ×αβ

Substituting the value of  α+β and  αβ from equation (1) we get,

α⁴ + β⁴ = ( $\frac{-b}{a}$ )⁴ + 2( $\frac{c}{a}$)² - 4( $\frac{-b}{a}$ )² × $\frac{c}{a}$

= $\frac{b^4}{a^4}$ + $\frac{2c^2}{a^2}$-$\frac{4b^2c}{a^3}$

= $\frac{b^4+2a^2c^2 - 4ab^2c}{a^4}$

∴α⁴ + β⁴ =  $\frac{b^4+2a^2c^2 - 4ab^2c}{a^4}$

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