Question

4. If 2 and - 3 are the zeroes of the quadratic polynomial x + (a +1)x + b. then find the
values of a and b.
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Answer 1

Error :- It will be x² + ( a + 1 )x + b instead of x + ( a + 1 )x + b .

Answer :-

→ a = 0 and b = -6 .

step-by-step explanation :-

We have ,

→ A quadratic polynomial = x² + ( a + 1 )x + b .

Here, A = 1 , B = ( a + 1 ) and C = b .

→ 2 and -3 are zeros of this polynomial .

Now,

Let α = 2 and β = -3 .

We know that ,

$\because$ α + β = - B/A .

⇒ 2 + ( -3 ) = -( a + 1 )/1 .

⇒ -1 = - a - 1 .

⇒ a = - 1 + 1 .

∴ a = 0 .

And,

$\because$ α × β = C/A .

⇒ 2 × (-3) = b/1 .

⇒ -6 = b .

∴ b = - 6 .

Hence, it is solved .

Answer 2

Given quadratic polynomial = x² + ( a + 1 )x + b .

A = 1 , B = ( a + 1 ) and C = b .

Let α = 2 and β = -3 .

∵ α + β = - B/A .

⇒ 2 + ( -3 ) = -( a + 1 )/1 .

⇒ -1 = - a - 1 .

⇒ a = - 1 + 1 .

∴ a = 0 .

∵ α × β = C/A .

⇒ 2 × (-3) = b/1 .

⇒ -6 = b .

∴ b = - 6 .