Question

the zeroes of quadratic polynomial are x²+25x+156 are​

Answer 1

Answer :-

Method I :-

By middle term splitting method :-

→ x² + 25x + 156

→ x² + 12x + 13x + 156

→ x ( x + 12 ) + 13 ( x + 12 )

→ ( x + 12 ) ( x + 13 )

→ x + 12 = 0

• x = -12

→ x + 13 = 0

• x = -13

Method II :-

Using quadratic formula :-

$\implies\sf x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

For the given equation :-

• a = 1
• b = 25
• c = 156

$\implies\sf x = \dfrac{-25 \pm \sqrt{25^2 - 4 \times 156 \times 1}}{2 \times 1}$

$\implies\sf x = \dfrac{-25 \pm \sqrt{625 - 624}}{2}$

$\implies\sf x = \dfrac{-25 \pm \sqrt1}{2}$

$\implies\sf x = \dfrac{-25 \pm 1}{2}$

$\implies\sf x = \dfrac{-25 + 1}{2} , \dfrac{-25 - 1}{2}$

$\implies\sf x = \dfrac{-24}{2} , \dfrac{-26}{2}$

$\implies\sf x = - 12 , -13$

Zeros of the polynomial = -12 , -13