Question

If the sum of the zeroes of the polynomial P (x) = (k²-14)x²-2x-12 is 1,find the value of k

Answer 1

Answer:

The values of k are -4 and 4

Step-by-step explanation:

The given polynomial is $(k^2-14)x^2-2x-12$

The sum of zeros of a quadratic equation is $-\frac{b}{a}$

Hence, the sum of the zeros of the given quadratic is

$-\frac{-2}{(k^2-14}\\\\\frac{2}{(k^2-14}$

Now, from the given directions, this expression should be equal to 1.

$\frac{2}{(k^2-14}=1\\\\k^2-14=2\\\\k^2=16\\\\k=\pm\sqrt{16}\\\\k=\pm4$

Answer 2

Answer:

Concept:

The negative of the coefficient of x by the coefficient of $x^{2}$ is equal to the sum of the zeroes in a quadratic polynomial. The constant term divided by the coefficient of $x^{2}$ is equal to the product of the zeroes. A zero polynomial is a polynomial with the value 0.

Step-by-step explanation:

Given:

If the sum of the zeroes of the polynomial P (x)  = $\left(K^{2}-14\right) x^{2}-2 x-12$ is 1

Find:

The value of K

Solution:

               $\left(K^{2}-14\right) x^{2}-2 x-12$

Therefore, $a=\left(\mathrm{K}^{2}-14\right), b=-2, c=-12$

The sum of zeroes of a quadratic polynomial $=-(-2) / k^{2}-14$

                          $2 / \mathrm{k}^{2}-14=1\\\\2=1\left(k^{2}-14\right)\\\\2=\mathrm{k}^{2}-14 \mathrm{K}^{2}=+2+14$

                                $K^{2}=16\\K=\sqrt{1} 6$

                                K = ± 4

If the sum of the zeroes of the polynomial P (x) = (k²-14)x²-2x-12 is 1,the value of k is  ± 4

 

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