Question

If $(\alpha +\surd \beta )$ and $(\alpha -\surd \beta )$ are the roots of the equation $x^2+px+q=0$, where $\alpha ,\beta ,p$ and $q$ are real, then the roots of the equation $(p^2-4q)(p^2{x}^2+4px)-16q=0$ are

• A. $\frac{1}{ɑ}+\frac{1}{\surd \beta }$ and $\frac{1}{ɑ}-\frac{1}{\surd \beta }$
• B. $\frac{1}{\surd ɑ}+\frac{1}{\beta }$ and $\frac{1}{\surd ɑ}-\frac{1}{\beta }$
• C. $\frac{1}{\surd ɑ}+\frac{1}{\surd \beta }$ and $\frac{1}{\surd ɑ}-\frac{1}{\surd \beta }$
• D. $\surd ɑ+\surd \beta$ and $\surd ɑ-\surd \beta$

Answer 1

The correct option is A

$\frac{1}{ɑ}+\frac{1}{\surd \beta }$ and $\frac{1}{ɑ}-\frac{1}{\surd \beta }$

Explanation for the correct option:

Step 1. Find the roots of $(p^2-4q)(p^2{x}^2+4px)-16q=0$:

Given, $(\alpha +\surd \beta )$ and $(\alpha -\surd \beta )$ are the roots of the equation $x^2+px+q=0$

$\therefore$Sum of the roots $=-p$

$\Rightarrow$$(\alpha +\surd \beta )+(\alpha –\surd \beta )=–p$

$\Rightarrow$ $2\alpha =–p$

$\Rightarrow$ $\alpha =\frac{–p}{2}$

and product of roots $=q$

$\Rightarrow$$(\alpha +\surd \beta )(\alpha –\surd \beta )=q$

$\Rightarrow$ ${\alpha }^2-\beta =q$

Step 2. Put the value of ${\alpha }^2$ in ${\alpha }^2–\beta =q$

$\frac{p^2}{4}–\beta =q$

$\Rightarrow$ $\beta =\frac{p^2–4q}{4}$

Step 3. put the values of $\beta$ and ${\alpha }^2$ in $(p^2-4q)(p^2{x}^2+4px)-16q=0$, we get

$(p^2-4q)(p^2{x}^2+4px)-16q=0$

$\Rightarrow$$4\beta (4{\alpha }^2{x}^2–8\alpha x)–16({\alpha }^2–\beta )=0$

$\Rightarrow$ $\beta (4{\alpha }^2{x}^2–8\alpha x)–4({\alpha }^2–\beta )=0$

$\Rightarrow$ ${\alpha }^2\beta x^2-2\alpha \beta x+\beta =\alpha$

$\Rightarrow$ ${\left(\alpha \surd \beta x–\surd \beta \right)}^2={\alpha }^2$

$\Rightarrow$ $\alpha \surd \beta x–\surd \beta =\pm \alpha$

$\Rightarrow$ $x=\frac{1}{\alpha }\pm \frac{1}{\beta }$

$\therefore$The required roots are $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{ɑ}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{\surd }\mathbf{\beta }}$ and $\frac{\mathbf{1}}{\mathbf{ɑ}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{1}}{\mathbf{\surd }\mathbf{\beta }}$

Hence, Option ‘A’ is Correct.