Question

If $\alpha \ne \beta$and $a^2=5\alpha -3$, ${\beta }^2=5\beta -3$, then the equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots is

• A. $3x^2+19x+3=0$
• B. $3x^2-19x+3=0$
• C. $3x^2-19x-3=0$
• D. $3x^2+19x+1=0$

Answer 1

The correct option is B

$3x^2-19x+3=0$

Explanation for the correct option:

Step 1. Find the required quadratic equation:

Given, $\alpha \ne \beta$, $a^2=5\alpha -3$ and ${\beta }^2=5\beta -3$

Change $\alpha ,\beta$ by $x$ in given equations, we get

$x^2-5x+3=0$

Here, $\alpha$and $\beta$are roots

$\Rightarrow$$\alpha +\beta =5,\alpha \beta =3$

Step 2. Find the $\alpha +\beta$ and $\alpha \beta$ of roots $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$:

$\begin{array}{rcl}\alpha +\beta & =& \frac{\alpha }{\beta }+\frac{\beta }{\alpha }\\ & =& \frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\\ & =& \frac{19}{3}\end{array}$

$\begin{array}{rcl}\alpha \beta & =& \frac{\alpha }{\beta }\times \frac{\beta }{\alpha }\\ & =& 1\end{array}$

Step 3. The required equation is $x^2–(\alpha +\beta )x+\alpha \beta =0$

$\Rightarrow$ $x^2–\left(\frac{19}{3}\right)x+1=0$

$\Rightarrow$ $\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{19}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence, Option ‘B’ is Correct.