Question

if one of the zeros of the quadratic polynomial (a-1)x2+ax+1 is -3, then find the value of a

Answer 1

(a-1)2 + ax + 1 ( zero of the polynomial=(-3) )
So, x=(-3)
=(a-1)2 + a(-3) +1=0
=2a-2+(-3a) +1=0
=2a-2-3a+1=0
= -a-1 =0
= -a = 1

Therefore a = (-1) //

Answer 2

Answer:-

a = 4/3

Given:-

p (x) = (a - 1) x² + ax + 1

-3 is one of the root of given quadratic polynomial.

To find :-

The value of a.

Solution:-

-3 is the root of the given polynomial I. e, when we put the value in equation it satisfies the equation as 0.

Now,

put the -3 in place of x,

→$( a-1) (-3) ^2 +a.(-3) +1 = 0$

→$(a -1) 9 -3a +1 = 0$

→$9a -9 -3a +1 = 0$

→$9a -3a -9 +1 = 0$

→$6a -8 =0$

→$6a = 8$

→$a = \dfrac{8}{6}$

→$a = \dfrac{4}{3}$

hence,

The value of a will be 4/3.