Question

Find quadratic polynomial whose zeroes are $3,-3.$

Answer 1

Step 1. Defining the root of quadratic polynomial.

We know that for a quadratic equation $a{x}^2+bx+c=0$ if $\alpha$ and $\beta$ are the two roots of the quadratic equation then

$\alpha +\beta =-\frac{b}{a}$and $\alpha \beta =\frac{c}{a}$

Step 2. Finding the value of $\frac{b}{a}$

Here $\alpha =3$ and $\beta =-3$

$\therefore \alpha +\beta =0$and $\alpha \beta =-9$

So $\frac{b}{a}=0$

Step 3. Finding the value of $\frac{c}{a}$

Here $\alpha =3$ and $\beta =-3$

$\therefore \alpha \beta =-9$

So $\frac{c}{a}=-9$

Step 4. Comparison of quadratic polynomial.

$$\begin{gathered} a{x}^2+bx+c=0 \\ \Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0 \\ \Rightarrow x^2+0.x-9=0[\because \frac{b}{a}=0,\frac{c}{a}=-9] \\ \Rightarrow x^2-9=0 \end{gathered}$$

Hence, the polynomial is $x^2-9.$