Question

if the zeroes of the polynomial p(x)= x^2+(a+1)x+b are 2 and -3, then find the value of (a+b)​

Answer 1

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

if the zeroes of the polynomial p(x)= x^2+(a+1)x+b are 2 and -3, then find the value of (a+b)

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

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

• polynomial , p(x) = x² + (a+1)x + b
• 2 and 3 are zeroes of this polynomial

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

• Value of (a+b)

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

We know,

If 2 and 3 are zeroes of this equation ,

So , if we keep x = 2,3 in this equation they this equation .

Case(1):-

• Keep x = 2 in this equation

➠ p(2) = 2² + (a+1).2 + b = 0

➠ 2a + b = - 4 - 2

➠ 2a + b = -6 .....................(1)

Case(2):-

• keep x = 3 in this equation

➠ p(3) = 3² + (a+1).3 + b = 0

➠ 3a + b = -9 - 3

➠3a + b = -12 ......................(2)

Multiply by 3 in equ(1) and 2 in equ(2)

• 6a + 3b = -18
• 6a + 2b = - 24

______________Subtract it's

➠ ( 3b - 2b) = (-18+24)

➠ b = 6

Keep value of b in equ(1),

➠ 2a + 6 = -6

➠ 2a = -6 - 6

➠ 2a = -12

➠ a = -12/2

➠ a = -6

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

• Value of a = -6
• Value of b = 6

Now, calculate :-

➠ (a + b)

Keep value of a and b

➠ ( -6 + 6)

➠ 0 ( Ans.)

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