If one of the zeroes of the quadratic polynomial $k x^{2} + k x - 16$ is $3$ , then the value of $k$ is

\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 B $\frac{4}{3}$
Given polynomial: $k x^{2} + k x - 16$
$3$ is the root of the equation, implies that it satisfies the above polynomial.
$∴ k (3)^{2} + 3 k - 16 = 0$
$9 k + 3 k - 16 = 0$
$12 k = 16$
$∴ k = \frac{4}{3}$

Hence, option A is the correct answer.

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Similar questions

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

(A) \frac{4}{3}

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

(C) \frac{2}{3}

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

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

α

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

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If one zero of the quadratic polynomial $k x^{2} + 3 x + k$ is 2 then find the value of k.

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