Question

if -2 and 3 are the zeros of the quadratic polynomial x²+(a+1)X+b,.then find the value of a and b​

Answer 1

Topic :-

Quadratic equations

Answer :-

• a = -2
• b = -6

STEP BY STEP EXPLANATION:-

Given :-

• 2, -3 are the zeros of Quadratic equation  x²+(a+1)x+b

To find :-

• Value of a, b

Solution :-

As , they given 2, -3 are the zeros of the Quadratic equation . Since , If we substitute the value of the zeros  in place of x it should be equal to zero . Then after substitution we get the two equations in the variable of a, b . After solving this two equations we can get the value of a, b .

Case -1 :-

At x = -2

x²+(a+1)x+ b

(-2)² +(a +1) (-2) +b = 0

4 -2a -2 +b =0

2 -2a + b =0 [eq 1]

Case -2 :-

At x = 3

x²+(a+1)x+ b =0

(3)² +(a+1) 3+b=0

9+ 3a+3+b =0

12+3a+b =0 [eq 2]

Subtracting the both equations

2 -2a +b -[ 12+3a+b] = 0

2-2a +b -12-3a - b =0

-10 -5a = 0

-5a = 10

a = -10/5

a = -2

Substitute value of a in equation -1

2 -2a +b = 0

2- 2(-2) +b = 0

2 + 4 +b =0

b +6 =0

b = -6

So, the values of a, b are -2, -6

Verification :-

We got the values of a, b Since If we substitute the values of a, b and x it should be equal to 0 .

x²+(a+1)x+ b =0

At ,

• x = -2
• a = -2
• b = -6

(-2)² +(-2+1)(-2) -6 =0

4 +(-1)(-2)-6 =0

4+2-6 =0

6-6 =0

0=0

L.H.S = R.H.S

Case -1 Verified !

At ,

• x = 3
• a = -2
• b = -6

x²+(a+1)x+ b =0  

(3)² +(-2+1)(3) -6 =0

9 +(-1)(3) - 6 =0

9-3-6=0

9-9 =0

0=0

L.H.S = R.H.S

Case-2 Verified !