Question

If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and –3, then ______.

Options

• a = –7, b = –1

• a = 5, b = –1

• a = 2, b = –6

• a = 0, b = –6

Answer 1

If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and –3, then a = 0, b = –6.

Explanation:

Given, x² + (a + 1) x + b = 0

Sum of the zeroes = α + β = `-((a + 1))/1` = – a – 1

and product of the zeroes = αβ = `b/1` = b

Here, α = 2, β = –3

∴ 2 – 3 = –a – 1

`\implies` a = 0

and (2) (–3) = b

`\implies` b = –6

∴ a = 0 and b = –6

Answer 2

If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and –3, then a = 0, b = – 6.

Explanation:

According to the question,

x² + (a + 1)x + b

Given that, the zeroes of the polynomial = 2 and –3,

When x = 2

2² + (a + 1)(2) + b = 0

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = –6  ........(1)

When x = –3,

(–3)² + (a + 1)(–3) + b = 0

9 – 3a – 3 + b = 0

6 – 3a + b = 0

–3a + b = –6   ......(2)

Subtracting equation (2) from (1)

2a + b – (–3a + b) = –6 – (–6)

2a + b + 3a – b = –6 + 6

5a = 0

a = 0

Substituting the value of ‘a’ in equation (1), we get,

2a + b = –6

2(0) + b = –6

b = –6