Question

If the zeros of the quadratic polynomial x^2+(a+1)x+ b are 2 and -3 then find a and b?

Answer 1

EXPLANATION.

Zeroes if the quadratic polynomial

x^2 + ( a + 1 )x + b are 2 and -3

To find a and b.

put the first zeroes in polynomial = 2

( 2)^2 + ( a + 1 )2 + b = 0

4 + 2a + 2 + b = 0

2a + b + 6 = 0 .......(1)

put the second zeroes in polynomial = -3

( -3)^2 + ( a + 1 ) (-3) + b = 0

9 + ( -3a - 3 ) + b = 0

9 - 3a - 3 + b = 0

- 3a + 6 + b = 0 ...... (2)

From equation (1) and (2)

multiply equation (1) by 3

multiply equation (2) by 2

we get,

6a + 3b + 18 = 0

-6a + 2b + 12 = 0

we get,

5b + 30 = 0

b = -6

put value of b = -6 in equation (1)

we get,

2a - 6 + 6 = 0

a = 0

Therefore,

value of a = 0 and b = -6

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• A quadratic polynomial x² + (a + 1)x + b

• The zeroes of the quadratic polynomial are 2 and -3

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The value of a and b in x² + (a + 1)x + b

${\green{\underline{\underline{\bold{Solution:-}}}}}$

When the zeroes of a quadratic polynomial are given the quadratic polynomial is

→ x² - (sum of zeroes)x + product of zeroes

The zeroes are 2 and -3

Sum of zeroes = 2 + (-3) = -1

Product of zeroes = 2(-3) = -6

The quadratic equation is

→ x² - (-1) x + (-6)

→ x² + x - 6

Comparing x² + x - 6 with x² + (a + 1)x + b

→ a + 1 = 1

→ a = 1 - 1 = 0

→ b = -6

Hence:-

$\large\boxed{ \rm \purple {\rightarrow a = 0 \: \: and \: b \: = - 6 }}$