$If one of the zeros of quadratic polynomial (k - 1) x^{2} + k x + 1 is - 3,$
$then find the value of k .$

\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 $\frac{4}{3}$
$Given: p (x) = (k - 1) x^{2} + k x + 1 and - p (- 3) = 0$

$∴ (k - 1) (- 3)^{2} + k (- 3) + 1 = 0 = 9 (k - 1) - 3 k + 1 = 0 = 9 k - 9 - 3 k + 1 = 0 = 6 k - 8 = 0 = 6 k = 8 \Rightarrow k = \frac{8}{6} = \frac{4}{3}$

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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}

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

f (x) = a x^{2} + b x + c t h e n, f (α)

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

then find the value of k .

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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