Question

if alpha and beta are the zeroes of quadratic polynomial x^2+3x+6 find the value of alpha square +beta square

Answer 1

p(x) = x²+3x+6 =0

alpha and beta are two zeroes

Let alpha = x and beta = y {becoz these r not there in keyboard}

x+y = -b/a = -3
xy= c/a = 6

Now, x²+y² = (x+y)²-2xy

= (-3)²-2*6
= 9-12
= -3

Answer 2

Answer:

The value of $\alpha^2+\beta^2$ is -3.

Step-by-step explanation:

Given alpha and beta are the zeroes of quadratic polynomial $x^2+3x+6$ we have to find the value of $\alpha^2+\beta^2$

Polynomial is $p(x) = x^2+3x+6$

Comparing above equation with

$ax^2+bx+c$, we get a=1, b=3 and c=6

$\text{Sum of zeroes=}\alpha+\beta=\frac{-b}{a}=\frac{-3}{1}=-3$

$\text{Product of zeroes=}\alpha \beta=\frac{c}{a}=\frac{6}{1}=6$

Now,

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha \beta=(-3)^2-2(6)=-3$

Hence,  the value of $\alpha^2+\beta^2$ is -3.