Question

if alpha. and beta are zeros of x2 - 4x + 3 find alpha 4 beta 3 + alpha 3 beta 4

Answer 1

Answer:

The value of ${\alpha}^4{\beta}^3+{\alpha}^3{\beta}^4$ is 108

Step-by-step explanation:

Concept:

If $\alpha$ and $\beta$ are roots of $ax^2+bx+c=0$ then

sum of roots=$\alpha+\beta=\frac{-b}{a}$

product of roots=$\alpha.\beta=\frac{c}{a}$

Given:

$\alpha\:and\:\beta$ are roots of

$x^2-4x+3=0$

Then,

sum of roots=$\alpha+\beta=4$

product of roots=$\alpha.\beta=3$

Now,

${\alpha}^4{\beta}^3+{\alpha}^3{\beta}^4\\\\={\alpha}^3{\beta}^3(\alpha+\beta)\\\\=(\alpha\beta)^3(\alpha+\beta)\\\\=(3)^3(4) \\\\=27(4)\\\\=108$

Answer 2

Answer:

108

Step-by-step explanation:

if alpha. and beta are zeros of x2 - 4x + 3 find alpha 4 beta 3 + alpha 3 beta 4

α⁴β³ + α³β⁴

= α³β³(α+β)

=(αβ)³(α+β)

x² - 4x + 3

α+β = -(-4)/1 = 4

αβ = 3/1 = 3

α⁴β³ + α³β⁴

= (3)³(4)

= (27)(4)

= 108