If one of the zeroes of the quadratic polynomial $(k - 1) x^{2} + k x + 1$ is $(- 3)$ , then k equals to:

\frac{4}{3}

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- \frac{4}{3}

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\frac{2}{3}

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- \frac{2}{3}

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Solution

The correct option is C $\frac{4}{3}$
Given $:$ One of the zeros of quadratic polynomial $(k - 1) x^{2} + k x + 1$ is $- 3$
To find $:$ Value of $k$
Solution $:$ As $- 3$ is one of the zeros of quadratic polynomial $(k - 1) x^{2} + k x + 1$
$\Rightarrow$ When we substitute $- 3$ in the polynomial , we get $0$

$⟹ (k - 1) (- 3)^{2} + k (- 3) + 1 = 0$
$⟹ 9 k - 9 - 3 k + 1 = 0$
$⟹ 6 k - 8 = 0$

or $k = \frac{8}{6} = \frac{4}{3}$

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Similar questions

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

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

(A) \frac{4}{3}

(B) \frac{- 4}{3}

(C) \frac{2}{3}

(D) \frac{- 2}{3}

If one zero of the quadratic polynomial $(k - 1) x^{2} + k x + 1$ is -4 then the value of k is

(a) $\frac{- 5}{4}$ (b) $\frac{5}{4}$ (c) $\frac{- 4}{3}$ (d) $\frac{4}{3}$

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

α

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If the sum of the zeros of the quadratic polynomial $k x^{2} + 2 x + 3 k$ is equal to the product of its zeros then k = ?

(a) $\frac{1}{3}$ (b) $\frac{- 1}{3}$ (c) $\frac{2}{3}$ (d) $\frac{- 2}{3}$

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