Question

If -2 and 3 are the zeros of the quadratic polynomial x2+(a+1)x+b then solution

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If -2 and 3 are the zeros of the quadratic polynomial x²+(a+1)x+b = 0, then solution ?

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Equation,x² + (a+1).x + b = 0--------(1)

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Value of "a" and " b"

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know, If p and q are zeroes of equation Ax² + Bx + C = 0 .

So, p and q satisfied this equation.

Since,

-2 and 3 are zeroes of equation (1),

It means that these value are satisfied of this equation,

Case(1):-

• when , x = -2.

keep value in equ(1),

➛ (-2)² + (a+1).(-2) + b = 0

➛ -2a -2 + 4 + b = 0

➛ 2a - b = 2 --------------(2)

Case(2):-

• when, x = 3

➛ 3² + (a+1).3 + b = 0

➛ 3a + b = - 9 - 3

➛ 3a + b = -12 --------------(3)

Add equ(2) and equ(3),

➛ (2a + 3a ) = (2 - 12)

➛ 5a = -10

➛ a = -10/5

➛ a = -2

keep value in equ(2),

➛ 2*(-2) - b = 2

➛ b = -2 - 4

➛ b = -6

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Value of a = -2
• Value of b = -6

____________________________

Answer 2

Answer: a = -2, 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) = (-2)² + (a + 1)(-2) + b = 0

4 - 2a - 2 + b = 0

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

p(3) = 3² + (a + 1)3 + b = 0

9 + 3a + 3 + b = 0

3a + b + 12 = 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 + 2 = 0

b = 2a - 2

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

Second Equation: 3a + b + 12 = 0

3a + (2a - 2) + 12 = 0

3a + 2a - 2 + 12 = 0

5a + 10 = 0

5a = -10

a = -2

Now, find the value of b.

b = 2a - 2

b = 2(-2) - 2

b = - 4 - 2

b = -6