Question

If the zeroes of the quadratic polynomial 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

Answer 1

Answer:

$\large\boxed{\star\:\:(d.)\:a=0\:,\:b=(-6)\:\:\star}$

Step-by-step explanation:

✢ Given:-

    The roots of Quadratic Equation x²+(a+1)x+b=0 are 2 and (-3).

✢ To Find:-

    The Values of a and b.

✢ Solution:-

p(x)= x²+(a+1)x+b=0

★ when x=2,

⇒p(2)=(2)²+(a+1)(2)+b=0

⇒ 4+2a+2+b=0

⇒ 2a+b=(-6)--------------(1)

★ when x=(-3),

⇒ p(-3)=(-3)²+(a+1)(-3)+b=0

⇒ 9-3a-3+b=0

⇒3a-b=6------------------(2)

★ Adding Equation(1) and Equation(2):-

⇒2a+3a+b-b=(-6)+6

⇒ 5a=0

$\implies\:a=\frac{0}{5}$

$\implies \boxed{a=0}$

★ Now  putting the value a=0 in Equation(2):-

⇒ 3(0)-b=6

⇒ 0-b=6

$\implies \boxed{b=(-6)}$