Question

. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then find a and b

Answer 1

Answer:

alpha+beta= -8 = -b/a

alpha beta = 12 = c/a

by the formula

ax^+bx+c =0

a=1;b=8;c=12

so the quadratic equation is 1x^+8+12=0

soving quadratic equations

1x^+8x+12=0(by middle term splitting method)

firstly multiply last term and first term that is 12×1x^=12x^

now when we factorise 12x^ by 6x ×2x we get 12x^ and when we add the multiples 6 and 2 we should get 8.So now the equation is 1x^+6x+2x+12=0

now take out common term

x( x+6)+ 2( x+6)=0

(x+2) (x+6)

x+2=0. x+6=0

x= -2 x= -6

So the zeores of polynomial is

-2. and -6

hope this is helpful.

mark me in brilliant list

Answer 2

Answer: a = 0, b = -6

Step-by-step explanation:

p(x) = x² + (a + 1)x + b

Zeroes of this polynomial are 2 and (-3). This means if we put 2 and (-3) as x in the above equation, we should get 0.

p(2) = 2² + (a + 1)2 + b = 0

4 + 2a + 2 + b = 0

2a + b + 6 = 0 (First Equation)

p(-3) = (-3)² + (a + 1)(-3) + b = 0

9 - 3a - 3 + b = 0

-3a + b + 6 = 0 (Second Equation)

Now, we have two equations and two variables. So we can find the values of a and b.

First Equation: 2a + b + 6 = 0

b = -2a - 6

Put the value of b obtained here in the second equation.

Second Equation: -3a + b + 6 = 0

-3a + (-2a - 6) + 6 = 0

-3a - 2a - 6 + 6 = 0

-5a = 0

a = 0

Now, find the value of b.

b = -2a - 6

b = -2(0) - 6

b = 0 - 6

b = -6