Question

if alpha,beta are the zeroes of quadratic polynomial x2-7x+10,find the value of alpha^3+beta^3​

Answer 1

Answer:

133

Step-by-step explanation:

Given :

If α & β are the zeroes of the polynomial, x² - 7x + 10 = 0 , then

Find α³ + β³

Solution :

We know that,

⇒ (a³ + b³) = (a + b) (a² - ab + b²)

_

We also know that,

For a quadratic equation of the form :

ax² + bx + c = 0,.

Here, a = 1 , b = -7 , c = 10

Sum of the zeroes = α + β = $\frac{-b}{a}$

Product of the zeroes = αβ = $\frac{c}{a}$

⇒ (a³ + b³) = (a + b) (a² - ab + b²)

= (a + b) (a² + 2ab - 3ab + b²)

= (a + b) [ (a² + 2ab + b²)  - 3ab ]

⇒ (a + b) [(a + b)² - 3ab]

Hence,

⇒ α³ + β³ = $( \frac{-b}{a} )[(\frac{-b}{a} )^2 - 3(\frac{c}{a})]$

⇒$(\frac{-(-7)}{1})[(\frac{-(-7)}{1})^2 - 3(\frac{10}{1})] = (7)[(7)^2 - 3(10)]$

⇒7(49 - 30) = 7 × 19 = 133

∴ α³ + β³ = 133

Answer 2

Answer:

133

Steps

x2--7x+10

x2-5x-2x+10

x(x-5)-2(x-5)

(x-2)(x-5)

x=2 or 5

let alpha =2 and beta= 5

now

alpha^3 + beta^3

(2)^3 + (5)^3

8 + 125

133