Question

if alpha and beta are the zeros of the quadratic polynomial 9x²-1,then find the value of alpha² and beta²

Answer 1

9x²-1=0
(3x)²-1²=0
Using a²-b²=(a+b)(a-b)
(3x+1)(3x-1)=0
X= -1/3 or 1/3
X²= (1/3)²
=1/9

Answer 2

Hey There,
Here, the polynomial is $9x^2 + 0.x -1$. On comparing it with the standard form $ax^2+bx+c$ we see that a=9, b=0, c= -1 .
Now, $\alpha$ and $\beta$ are the zeros of the quadratic polynomial.  
So, $\alpha+\beta=\frac{-b}{a}=\frac{-0}{9} = 0$ 
Also, $\alpha\beta=\frac{c}{a}=\frac{-1}{9}$
Now, $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta \\ \\ \implies 0^2 = \alpha^2+\beta^2 + 2\frac{-1}{9} \\ \\ \implies \boxed{\alpha^2+\beta^2 = \frac{2}{9}}$
_____________________________
There is also an alternative method.
$9x^2-1=0 \\ \\ \implies (3x-1)(3x+1) = 0 \\ \\ \implies x=\pm\frac{1}{3} \\ \\ \implies \alpha = \frac{1}{3} \, and \, \beta=\frac{-1}{3} \\ \\ \implies \alpha^2+\beta^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{-1}{3}\right)^2 \\ \\ \implies \boxed{\alpha^2+\beta^2=\frac{2}{9}}$ 

Hope it helps
Purva
Brainly Community