If one of the zeroes of the quadratic polynomial $(k - 1) x^{2} + k x + 1$ 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 A
$\frac{4}{3}$

Find the value of $k$ .
Since, the roots of a quadratic polynomial satisfy the polynomial.
$\Rightarrow (k - 1) {(- 3)}^{2} + k (- 3) + 1 = 0 \Rightarrow (k - 1) 9 - 3 k + 1 = 0 \Rightarrow 9 k - 9 - 3 k + 1 = 0 \Rightarrow 6 k = 8 \Rightarrow k = \frac{4}{3}$
Hence, $Option (A)$ is correct.

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

if one zero of the quadratic polynomial (x-1)(x^2+kx+1) is -3 then value of k is

(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 of the zeros of quadratic polynomial (k - 1) x^{2} + k x + 1 is - 3,

then find the value of k .

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