Question

If one zero of the quadratic polynomial $f\left( x \right)=4{{x}^{2}}-8kx-9$ is negative of the other, then find the value of k.

Answer 1

Hint:In question given, one zero of the quadratic polynomial is negative of the other.Let consider zeroes of the given quadratic equation as $\alpha $ and $-\alpha $ and use the following relation among zeroes of any quadratic equation with its coefficients.
For any quadratic $a{{x}^{2}}+bx+c=0$ , we have
Sum of zeroes = $\dfrac{-b}{a}$
Product of zeroes = $\dfrac{c}{a}$

Complete step-by-step answer:
As we know the relation among zeroes and coefficients of any quadratic equation $a{{x}^{2}}+bx+c=0$ is given as
Sum of zeroes = $\dfrac{-b}{a}$ …………………(i)
Product of zeroes = $\dfrac{c}{a}$ ………………..(ii)
Now, coming to the question it is given that one zero of the given quadratic $f\left( x \right)=4{{x}^{2}}-8kx-9$ is negative of the other zero.
So, we have
$f\left( x \right)=4{{x}^{2}}-8kx-9.....................\left( iii \right)$
So, let us suppose the zeroes of the quadratic equation given in equation (iii) are $'\alpha '$ and $'-\alpha '$ .
Now, on comparing the equation (iii) with the general equation of a quadratic i.e. $a{{x}^{2}}+bx+c=0$ , we get
$a=4,b=-8k,c=-9$
So, using equations (i) and (ii), we get
$\begin{align}
  & \alpha -\alpha =\dfrac{-\left( -8k \right)}{4}=2k \\
 & \Rightarrow 0=2k \\
 & \Rightarrow k=0 \\
\end{align}$
So, the value of k is given as 0.

Note: Another approach for the question would be that we can put roots $\alpha $ and $-\alpha $ to the given equation as well.
So, on putting $\alpha $ to $f\left( x \right)$ , we get
$4{{\alpha }^{2}}-8k\alpha -9=0$
Similarly, put $-\alpha $ to $f\left( x \right)$ , we get
$4{{\alpha }^{2}}+8k\alpha -9=0$
On subtraction, we get k = 0.
So, it can be another approach. One may go for calculating the roots of the quadratic with the help of quadratic formula, given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for quadratic $a{{x}^{2}}+bx+c=0$
And hence, we can add the roots and equate them to 0. So, it can be another approach but a little more complex than given in the solution.