Question

4. If the zeroes of the quadratic
polynomial x2 + (a + 1) x + b are
2 and -3, then
O
(a) a = -7, b = -1
O (b) a = 5, b = -1
O (c) a = 2, b = -6
0 (, b
(d) a - 0, b = -6​

Answer 1

Step-by-step explanation:

substitute x=2 in x2+(a+1)x+b=0

4+(a+1)×2+b=0

2a+b= -6

substitute x=(-3) in the equation

9+(a+1)×(-3)+b=0

3a+b= -12

solving 2a+b=-6 and (-3a)+b= -6 we get

a=0 and b= -6 so ans is option (d)

Answer 2

$\huge\sf\underline\purple{Explanation:-}$

Correct Question :-

If the zeroes of the quadratic

equation x²+ (a + 1) x + b are

2 and -3, then

(a) a = -7, b = -1

(b) a = 5, b = -1

(c) a = 2, b = -6

(d) a = 0, b = - 6

Solution:-

So,Given x²+(a+1)x + b =0 roots are 2, -3

So, Hence roots are given So substuite in equation to get values of a,b

So Substuite x=2

(2)² + ( a + 1) 2 + b =0

4+2a+2+b =0

6+2a+b=0 ________eq1

So substuite x = -3

(-3)²+(a+1)-3+b =0

9-3a-3+b =0

6-3a +b =0 _______eq2

Subtracting eq 1 from 2

6 + 2a + b = 0

6 - 3a + b =0

___________

5a =0

a = 0

Sub value of a in eq 1

6+2a+b=0

6 + b =0

b = -6.

Hence your answer is option d

a = 0 b = -6