Question

If $\alpha ,\beta$ are the zeros of the polynomial $a{x}^2+bx+c,$$a\ne 0$, then find the quadratic polynomial whose zeros are ${\alpha }^{2}and{\beta }^2$

Answer 1

Zeroes of a polynomial:

The zeros of a polynomial $f\left(x\right)$ are all the values of $x$ that make the value of the polynomial equal to zero.

Example:

$$\begin{gathered} f\left(x\right)=3x^2-2x-1 \\ At,x=1, \\ f\left(1\right)=3\times 1^2-2\times 1-1 \\ =3-2-1 \\ =0 \end{gathered}$$

Then, $1$ is a zero of $f\left(x\right)$.

Given Data:

Given that, $\alpha ,\beta arethezerosofa{x}^2+bx+c$

Formula:

Let, $\alpha ,\beta$ are the zeroes of the quadratic polynomial $a{x}^2+bx+c$

Then,

$$\begin{gathered} \alpha +\beta =-\frac{b}{a}...\left(i\right) \\ \alpha \beta =\frac{c}{a}...\left(ii\right) \end{gathered}$$

Calculation:

Squaring (i) and (ii) on both sides,

We get, $$\begin{gathered} {(\alpha +\beta )}^2={(-\frac{b}{a})}^2 \\ or{\alpha }^2+{\beta }^2+2\alpha \beta =\frac{b^2}{a^2} \\ or{\alpha }^2+{\beta }^2=\frac{b^2}{a^2}-\frac{2c}{a}[\sin ce,\alpha \beta =\frac{c}{a}] \\ or{\alpha }^2+{\beta }^2=\frac{(b^2-2ac)}{a^2} \end{gathered}$$

and, $$\begin{gathered} {\left(\alpha \beta \right)}^2=\frac{c^2}{a^2} \\ or,{\alpha }^2{\beta }^2=\frac{c^2}{a^2} \end{gathered}$$

Now, The quadratic polynomial whose zeroes are ${\alpha }^{2}and{\beta }^2$

$x^2-(sumofthezeros)x+(productofthezeros)$

$$\begin{gathered} x^2-\frac{(b^2-2ac)}{a^2}x+\frac{c^2}{a^2} \\ =\frac{1}{a^2}[a^2{x}^2-(b^2-2ac)x+c^2] \end{gathered}$$

Conclusion:

The quadratic polynomial whose zeros are ${\alpha }^{2}and{\beta }^2$ is$k[a^2{x}^2-(b^2-2ac)x+c^2]$, Where $k$ is a non-zero constant.

Final answer:

Therefore, the required polynomial is $k[a^2{x}^2-(b^2-2ac)x+c^2]$, Where $k$ is a non-zero constant.