Question

If 2 and -3 are the zeroes of the quadratic polynomial
x2 + (a + 1)x + b; then find the values of

a

Answer 1

Answer:

a = 5

Step-by-step explanation:

given that the two roots are 2 and -3

we know that ,

$\alpha \beta = - b \div a$

$2 \times - 3 = - (a + 1 ) \div 1$

$- 6 = - (a + 1)$

$a + 1 = 6$

$a = 6 - 1$

$a = 5$

Answer 2

Given that:

• Quadratic polynomial: x² + (a + 1)x + b
• Two zeroes are 2 and - 3.

To Find:

• The values of a.

Solution:

Sum of zeroes = - (a + 1)/1

⟶ 2 + (- 3) = - a - 1

⟶ 2 - 3 = - a - 1

⟶ - 1 = - a - 1

⟶ a = - 1 + 1

⟶ a = 0

∴ The value of a = 0

Product of zeroes = b/1

⟶ 2 × (- 3) = b

⟶ - 6 = b

⟶ b = - 6

∴ The value of b = - 6