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

\frac{4}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- \frac{4}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{2}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

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

Suggest Corrections

214

Similar questions

(k - 1) x^{2} + k x + 1 i s - 3

f (x) = {\begin{matrix} 2 a^{2} x - 1 & ; & x \leq 1 x^{2} + x + b & ; & x > 1 \end{matrix}

a, b \in R

2 a^{2} + b

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

(- 3)

k

k x + {sin}^{- 1} (x^{2} - 8 x + 17) + {cos}^{- 1} (x^{2} - 8 x + 17) = \frac{9 π}{2}

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

(A) \frac{4}{3}

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

(C) \frac{2}{3}

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

Join BYJU'S Learning Program