Question

If the roots of equation $a^2{x}^2+b^{2}x+c^2=0$ are the squares of the roots of the equation $a{x}^2+bx+c=0$ then $a,b,c$ are in _____$(a,b,c,\in R-\{0\left\}\right)$

• A. G. P.
• B. H.P.
• C. A.P.
• D. None of these

Answer 1

The correct option is A

G. P.

Given: Here roots of $a^2{x}^2+b^{2}x+c^2=0$ are the squares of the roots of $a{x}^2+bx+c=0$

Step-1 Find the sum and product of the roots of c

Let $\alpha and\beta$ are the roots of $a{x}^2+bx+c$.

$$\begin{gathered} \alpha +\beta =\frac{-b}{a}...\left(1\right) \\ \Rightarrow \alpha \beta =\frac{c}{a}...\left(2\right) \end{gathered}$$

Let ${\alpha }_{1}and{\beta }_1$ are the roots of $a^2{x}^2+b^{2}x+c^2$

$$\begin{gathered} {\alpha }_1+{\beta }_1=\frac{-b^2}{a^2}...\left(3\right) \\ \Rightarrow {\alpha }_1{\beta }_1=\frac{c^2}{a^2}...\left(4\right) \end{gathered}$$

Step-2 Express the statement into mathematical form.

$$\begin{gathered} {\alpha }_1={\alpha }^2...\left(5\right) \\ {\beta }_2={\beta }^2...\left(6\right) \end{gathered}$$

Put equation number $\left(5\right)$ and $\left(6\right)$ in the equation number $\left(3\right)$ then, the equation $\left(3\right)$ becomes.

${\alpha }^2+{\beta }^2=\frac{-b^2}{a^2}$

Add and subtract $2\alpha \beta$ in the above equation.

${\alpha }^2+{\beta }^2+2\alpha \beta -2\alpha \beta =\frac{-b^2}{a^2}$

Use the identity ${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

${(\alpha +\beta )}^2-2\alpha \beta =\frac{-b^2}{a^2}...\left(7\right)$

Step-3 Eliminate $\alpha and\beta$ from the above equation ${(\alpha +\beta )}^2-2\alpha \beta =\frac{-b^2}{a^2}$

From equation number $\left(1\right),\left(2\right)and\left(7\right)$

$$\begin{gathered} {\left(\frac{-b}{a}\right)}^2-2\left(\frac{c}{a}\right)=\frac{-b^2}{a^2} \\ \Rightarrow \frac{b^2}{a^2}-\frac{2c}{a}=\frac{-b^2}{a^2} \end{gathered}$$

Adding both sides by $\frac{b^2}{a^2}$

$$\begin{gathered} \frac{b^2}{a^2}-\frac{2c}{a}+\frac{b^2}{a^2}=\frac{-b^2}{a^2}+\frac{b^2}{a^2} \\ \Rightarrow \frac{2b^2}{a^2}-\frac{2c}{a}=0 \end{gathered}$$

Adding both sides by $\frac{2c}{a}$

$$\begin{gathered} \frac{2b^2}{a^2}-\frac{2c}{a}+\frac{2c}{a}=\frac{2c}{a} \\ \Rightarrow \frac{2b^2}{a^2}=\frac{2c}{a} \\ \Rightarrow b^2=ac \end{gathered}$$

$b^2=ac$ is a condition for a, b, and c to be in G.P.

Therefore option (A) G. P. is the correct option.