Question

Find the zeroes of the quadratic polynomial $f\left( x \right)={{x}^{2}}-2x-8$ and verify the relationship between the zeroes and their coefficients.

Answer 1

Hint: First equate $f\left( x \right)=0$, and then by using the middle term factor method, find the roots. After that for verifying, use the formula; Sum of roots $=\dfrac{\text{- coefficient of }x}{\text{coefficient of }{{x}^{2}}}$ and Product of roots$\text{ =}\dfrac{\text{constant}}{\text{coefficient of }{{x}^{2}}}$ and get the answer to the given question.

Complete step-by-step solution -
In the question, we have to find the zeros of the polynomial $f\left( x \right)={{x}^{2}}-2x-8$ and then verify the relationship between the zeroes and the coefficients of the polynomial. As we have to find the zeros of the polynomial $f\left( x \right)$, we can say that we have to find the values of $x$ such that the value of $f\left( x \right)=0$. So, for $f\left( x \right)$ to be $0$, let us put it in an equation as follows,
$\begin{aligned} & f\left( x \right)={{x}^{2}}-2x-8 \\\ & \Rightarrow 0={{x}^{2}}-2x-8 \\\ \end{aligned}$
By reversing the sides of the equation, we get,
${{x}^{2}}-2x-8=0$
Now, by using middle term factor, we get,
${{x}^{2}}-4x+2x-8=0$
By further factorising, we get,
$\begin{aligned} & \left( x-4 \right)\left( x+2 \right)=0 \\\ & \Rightarrow x=4,\text{ }x=-2 \\\ \end{aligned}$
So, the values of $x$ for which $f\left( x \right)=0$ satisfies is $-2$ and $4$.
We know that the relationship between the coefficients and the roots that exist. The first one is,
Sum of roots $=\dfrac{-\text{ Coefficient of }x}{\text{Coefficient of }{{x}^{2}}}$.
We know that the roots are $-2$ and $4$. So, the sum of the roots is $-2+4=2$. We know the coefficient of $x$ is $-2$, and the coefficient of ${{x}^{2}}$ is $1$, so we get,
$\dfrac{-\text{ coefficient of }x}{\text{coefficient of }{{x}^{2}}}=\dfrac{-\left( -2 \right)}{1}=2$
Hence, the first relation is satisfied. Now, there exists another final relation which is,
Product of roots $=\dfrac{\text{constant }}{\text{coefficient of }{{x}^{2}}\text{ }}$.
We know that the roots are $-2$ and $4$. So, the product of the roots is $-2\times 4=-8$. We know that the constant term is $-8$, and the coefficient of ${{x}^{2}}$ is $1$, so we get,
$\dfrac{\text{constant }}{\text{coefficient of }{{x}^{2}}\text{ }}=\dfrac{-8}{1}=-8$
Hence, the final relation is also satisfied.
Therefore, the roots are $-2$ and $4$.

Note: The students should be careful while finding out the roots by using the middle term factor method and equating $f\left( x \right)=0$. They should also be careful while writing the relations of the roots to avoid any mistakes in proving the relations.