One of the zeroes of the polynomial $(k - 1) x^{2} + k x + 1 i s - 3$ . Find the other zero.

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Solution

$- 3$ is zero of polynomial $(k - 1) x^{2} + k x + 1$

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

$⟹ 9 k - 9 - 3 k + 1 = 0$

$⟹ 6 k = 8$

$= k = 4 / 3$

So equation is $(\frac{4}{3} - 1) x^{2} + \frac{4}{3} x + 1$

$= \frac{x^{2}}{3} + \frac{4 x}{3} + 1$

Or $x^{2} + 4 x + 3 = 0$

$x = - 1, - 3$

Other zero is $- 1$ .

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(k - 1) x^{2} + k x + 1

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(k - 1) x^{2} + k x + 1

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