Question

If the zeroes ofthe quadratic polynomial x^2+(a+1)x+bare 2 and -3

Answer 1

Step-by-step explanation:

Given -

• Zeroes of the polynomial p(x) = x² + (a + 1)x + b are 2 and -3

To Find -

• Value of a and b

As we know that :-

• α + β = -b/a

→ -3 + 2 = -(a + 1)/1

→ -1 = -a - 1

→ a = 0

And

• αβ = c/a

→ -3 × 2 = b/1

→ b = -6

Hence,

The value of a is 0 and b is -6

Verification :-

→ x² + (0 + 1)x + (-6)

→ x² + x - 6 = 0

→ x² - 2x + 3x - 6

→ x(x - 2) + 3(x - 2)

→ (x + 3)(x - 2)

Zeroes are -

→ x + 3 = 0 and x - 2 = 0

→ x = -3 and x = 2

Here the zeroes come same as given in the question.

It shows that our answer is absolutely correct.

Answer 2

$\large{\boxed{\underline{\underline{\bf{\red{Answer:-}}}}}}$

$\implies$ a = 0

$\implies$ b = -6

$\large\underline\mathrm{Given:-}$

• zeroes ofthe quadratic polynomial p(x) = x² + (a - 1)x + b
• α = 2
• β = -3

$\large\underline\mathrm{To \: find}$

• value of a and b.

$\large\underline\mathrm{We know that:-}$

$\implies$ α + β = -b/a

$\implies$ -3 + 2 = -(a + 1)/1

$\implies$ -1 = -a - 1

$\implies$ a = 0

$\large\underline\mathrm{and}$

$\implies$ βα = c/a

$\implies$ -3 × 2 = b/1

$\implies$ b = -6

$\large\underline\mathrm{hence,}$

The value of a is 0 and b is -6.

$\large\underline\mathrm{Verification \: of \: quadratic \: polynomial \: .}$

$\implies$ x² + (o - 1)x + -6

$\implies$ x² + x - 6 = 0

$\implies$ x² - 2x + 3x -

$\implies$ x(x - 2) + 3(x - 2)

$\implies$ x + 3 = 0

$\implies$ x = -3

$\implies$ x - 2 = 0

$\implies$ x = 2

$\large\underline\mathrm{Then,}$

$\large\underline\mathrm{The \: value \: of \: x² \: + \: (a - 1)x \: + \: b \: is \: x \: = \: -3 \: , \: 2}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$