Question

if alpha and beta are the zeroes of the polynomial x^2-6x+a,find alpha if beta is -2​

Answer 1

Step-by-step explanation:

Explanation

We are given a Quadratic polynomial and we are given one of the the roots ,

And we need to find the other root

.

Procedure

The given polynomial is x^2-6x+a=0

Now to find the value of a , we need to substitute value of beta in the equation

x^2-6x+a=0

4+12+a= 0

a= -16

Now we need to find alpha

alpha + beta = -b/a

alpha - 2 = 6

alpha = 8

Answer 2

$\boxed{\red{AnswEr}}$

$\alpha = 8$

★ Given :

• Quadratic equation, ${x}^{2} - 6x + a = 0$
• $\alpha \: and \: \beta \: \: are \: the \: zeroes \: \\ of \: the \: polynomial$
• $\beta = ( - 2)$

★ To Find :

$the \: value \: of \: \alpha$

★ Solution :

Method 1.

We know that, in any quadratic equation,(ax^2 + bx + c = 0) if alpha and beta are the zeroes of the polynomial then,

$\alpha + \beta = - \frac{b}{a} \\ \\ = > \alpha + ( - 2) = - \frac{( - 6)}{1} \\ \\ = > \alpha = 6 + 2 = 8$

Therefore, alpha = 8

Method 2.

Putting the value of Beta(-2) at the place of x in the given equation, we get,

${x}^{2} - 6x + a = 0 \\ \\ = > ( - {2}^{2} ) - 6 \times ( - 2) + a = 0 \\ \\ = > 4 + 12 + a = 0 \\ \\ = > a = - 16$

Now, putting the value of a in the equation we get,

${x}^{2} - 6x - 16 = 0 \\ \\ = > {x}^{2} - (8 - 2)x - 16 = 0 \\ \\ = > {x}^{2} - 8x + 2x - 16 = 0 \\ \\ = > x(x - 8) + 2(x - 8) = 0 \\ \\ = > (x - 8)(x + 2) = 0$

Therefore, x = 8 or (--2)

The value of Beta is --2 so the value of alpha is 8.