Question

If 2 and -3 are the zeros of the quadratic polynomial x2+(a+1)x+b than find the value of a

Answer 1

x² + (a+1)x + b

2 and -3 are zeroes of the polynomial,

put x = 2

» 2² + (a+1)(2) + b = 0

» 4 + 2a+2 + b = 0

» 6 + 2a+b = 0

» 2a+b = -6 -----(1)

put x = -3,

» (-3)² + (a+1)(-3) + b = 0

» 9 - 3a-3 + b = 0

» 6 - 3a+b = 0

» -3a+b = -6 -----(2)

(1) - (2)

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

2a+b+3a-b = -6+6

5a = 0

a = 0
________

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

Answer 2

Given :

2 and -3 are the zeros of the quadratic polynomial $x^2+(a+1)x+b$

To find :

a  = ?

Step-by-step explanation:

$x^2+(a+1)x+b$ is the quadratic polynomial.

2 and −3 are the zeros of the quadratic polynomial.

Thus, $2+(-3)= -(a+1) / 1$

=>$(a+1)/1 = 1$

=>a+1=1

=>a=0

Also, 2×(−3)=b

=>b=−6

Hence :

a = 0

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