Question

If $\alpha$ and $\beta$are the roots of the equation $x^2+ax+b=0,$ then $(1/{\alpha }^2)+(1/{\beta }^2)=$

• A. $\frac{(a^2-2b)}{b^2}$
• B. $\frac{(b^2-2a)}{b^2}$
• C. $\frac{(a^2+2b)}{b^2}$
• D. $\frac{(b^2+2a)}{b^2}$

Answer 1

The correct option is A

$\frac{(a^2-2b)}{b^2}$

Find the value of $\mathbf{(}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}{\mathit{\alpha }}^{\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}{\mathit{\beta }}^{\mathbf{2}}\mathbf{)}$:

Given that $\alpha$ and $\beta$are the roots of the equation $x^2+ax+b=0,$

$\Rightarrow (\alpha +\beta )=-a;\alpha .\beta =b$

Now, $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{({\beta }^2+{\alpha }^2)}{{(\alpha .\beta )}^2}$

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

Hence, the correct option is A.