Question

If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the

value of k.​

Answer 1

Answer:-

Given:

Sum of the zeroes of a polynomial (k² - 14)x - 2x - 12 = 1.

On comparing the polynomial with the standard form of a Quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = k² - 14

• b = - 2

• c = - 12

We know that,

Sum of the zeroes = - b/a

So, - ( - 2) / k² - 14 = 1

⟶ 2 = k² - 14

⟶ 2 + 14 = k²

⟶ 16 = k²

⟶ √16 = k

⟶ ± 4 = k

∴ The value of k is ± 4.

Answer 2

Step-by-step explanation:

Given:-

• A quadratic polynomial p(x) = (k² - 14) x² - 2x - 12

• The sum of zeroes of the polynomial is 1.

To Find:-

• The value of k.

Solution:-

For a quadratic polynomial ax² + bx + c

Sum of zeroes = $\sf \dfrac{-b}{c}$

In (k² - 14) x² - 2x - 12

• a = k² - 14

• b = -2

• c = -12

Sum of zeroes = 1

$\sf \implies \dfrac{ - ( - 2)}{ {k}^{2} - 14 } = 1$

$\sf \implies \dfrac{ 2}{ {k}^{2} - 14 } = 1$

$\sf \implies 2 = {k}^{2} - 14$

$\sf \implies {k}^{2} = 2 + 14$

$\sf \implies {k}^{2} = 16$

$\sf \implies k = \sqrt{16}$

$\sf \implies k = \pm4$

$\longrightarrow \underline{ \boxed{\sf \therefore k = \pm4}}$