Question

If one of the zeroes of the quadratic polynomial $\left( {k - 1} \right){x^2} + kx + 1$ is $\left( { - 3} \right)$ then k equals to:

Answer 1

Hint: In the question, a quadratic polynomial equation is given. For this, we will put the given value $\left( { - 3} \right)$ as the value of $x$ in the given equation, thus we will find the value of k and will get the correct answer.

Complete step by step solution: Given that:
One of the zeroes of the quadratic polynomial $\left( {k - 1} \right){x^2} + kx + 1$ is $\left( { - 3} \right)$
We have to find the value of k.
As $\left( { - 3} \right)$ is one of the zeroes of the quadratic polynomial $\left( {k - 1} \right){x^2} + kx + 1$
Put the value $\left( { - 3} \right)$ in the polynomial
We get
$\left( {k - 1} \right){\left( { - 3} \right)^2} + k\left( { - 3} \right) + 1 = 0$
We substitute $x = \left( { - 3} \right)$ in the given equation.
$
   \Rightarrow 9k - 9 - 3k + 1 = 0 \\
   \Rightarrow 6k - 8 = 0 \\
    \\
 $
From the above equation
$k = \dfrac{8}{6} = \dfrac{4}{3}$

From this, we got the value of k i.e. $\dfrac{4}{3}$

Note: If we think graphically then quadratic equations parabola on the coordinate plane. its intersection with x-axis represents its zeros. Also sign of ${x}^{2}$ tells us, whether the parabola is upside down or downside up.