Question

if 2 and -3 are the zeroes of the polynomial x^2+ax^2+bx-12 . find the values of a and b

Answer 1

ANSWER:

given

p(x)=x²+ax²+bx-12

now let us take x common

then

p(x)=(1+a)x²+bx-12

now also given that

$\alpha = 2 \\ \beta = - 3$

now

when do the product of zeroes we get the value of a

and then add zeroes we get the value of b

IDENTITIES USED:

$> sum \: of \: roots = \dfrac{ - b}{a}$

$> product \: of \: roots \: = \dfrac{c}{a}$

Answer 2

Given that 2 and -3 are the zeros of the above polynomial.

Let P(x) represent the given polynomial.

P(x)=x²+ax²+bx-12

We know that,

P(x)=P(-3)=P(2)=0

Now,

(2)²+a(2)²+b(2)-12=0 and (-3)²+a(-3)²+b(-3)-12=0

→4a+2b=12-4 and 9a-3b=12-9

→4a+2b=8 and 9a-3b=3

→2a+b=4..........[1]

and 3a-b=1...............[2]

Adding equations [1] and [2],

(2a+b)+(3a-b)=4+1

→5a=5

→a=1

Putting a=1 in [1],

2a+b=2

→2+b=2

→b=0

Hence,(a,b)=(1,0)