Question

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

Answer 1

Hint:We are provided the quadratic polynomial in the question. As we know the quadratic polynomial will have real and equal solutions, real and unequal solutions or imaginary complex conjugates as a solution. In our case we are given one solution -3 so the second root will also be real.

Complete step by step answer:
We have given the quadratic polynomial as $ (k - 1){x^2} + kx + 1$ and one of the zeros or we can call it as root is real and It Is -3. If we put -3 in the given equation then we will get the result as 0. So,
$ (k - 1){x^2} + kx + 1 = 0 \\
at\,x = - 3 \\
\Rightarrow (k - 1)({3^2}) + k(3) + 1 = 0 $
Further solving the equation we will get
$9(k - 1) + 3k + 1 = 0 \\
\Rightarrow 9k - 9 + 3k + 1 = 0 \\
\Rightarrow 12k = 8 \\
\therefore k=\dfrac{2}{3}$

Hence we get the value of $k$ as $\dfrac{2}{3}$.

Note:The quadratic equation will have different conditions for different roots. If $ a{x^2} + bx + c = 0$ is the quadratic equation then the $ D = \sqrt {\left( {{b^2} - 4ac} \right)} $ is called discriminant. If
(i) \[D < 0\] roots are imaginary.
(ii) \[D > 0\] roots are unequal and real.
(iii) \[D = 0\] roots are equal and real.