Question

5. If one of the zeroes of the quadratic polynomial (k-1) x2 + kx + 1 is --3, then find the value of k.
5. Find the quadratic polynomial
between the co-​

Answer 1

EXPLANATION.

One of the zeroes of the quadratic equation.

⇒ (k - 1)x² + kx + 1 is -3.

As we know that,

Put the value of x = -3 in equation, we get.

⇒ (k - 1)(-3)² + k(-3) + 1 = 0.

⇒ (k - 1)(9) - 3k + 1 = 0.

⇒ 9k - 9 - 3k + 1 = 0.

⇒ 9k - 3k - 9 + 1 = 0.

⇒ 6k - 8 = 0.

⇒ 6k = 8.

⇒ k = 8/6

⇒ k = 4/3.

                                                                                                                         

MORE INFORMATION.

Conjugate roots.

If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Answer:

Given :-

$\sf \: (k - 1) {x}^{2} + kx + 1 = - 3$

To Find :-

Value of k

Solution :-

$\sf \: (k - 1) {( - 3)}^{2} + k( - 3) + 1 = 0$

$\sf \: (k - 1)9 + - 3k + 1 = 0$

$\sf \: 9k - 9 - 3k + 1 = 0$

$\sf \: 6k - 9 - 1 = 0$

$\sf \: 6k - 8 = 0$

$\sf \: 6k = 0 + 8$

$\sf \: 6 k = 8$

$\sf \: k = \dfrac{8}{6}$

$\sf \: k = \dfrac{4}{3}$