Question

If $\alpha \ne \beta$ but ${\alpha }^2=2\alpha -3;{\beta }^2=2\beta -3$ then the equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ is

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

Answer 1

Answer: (B)

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

$\Rightarrow$ ${\alpha }^2=2\alpha -3$ and ${\beta }^2=2\beta -3$

$\Rightarrow$ ${\alpha }^2-2\alpha +3=0$ and ${\beta }^2-2\beta +3=0$

$\Rightarrow$ $\alpha =\frac{2\pm \sqrt{4-12}}{2}$ and $\beta =\frac{2\pm \sqrt{4-12}}{2}$

$\Rightarrow$ $\alpha =\frac{2\pm \sqrt{-8}}{2}$ and $\beta =\frac{2\pm \sqrt{-8}}{2}$

$\Rightarrow$ $\alpha =\frac{2\pm 2\sqrt{2}i}{2}$ and $\beta =\frac{2\pm 2\sqrt{2}i}{2}$

$\Rightarrow$ $\alpha =1\pm \sqrt{2}i$ and $\beta =1\pm \sqrt{2}i$

Since $\alpha \ne \beta$, we are taking $\alpha =1+\sqrt{2}i$ and $\beta =1-\sqrt{2}i$

Now,

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(1+\sqrt{2}i{)}^2+(1-\sqrt{2}i{)}^2}{(1+\sqrt{2}i)(1-\sqrt{2}i)}$

$=\frac{(1{)}^2+(\sqrt{2}i{)}^2+2\sqrt{2}i+(1{)}^2+(\sqrt{2}i{)}^2-2\sqrt{2}i}{(1{)}^2-(\sqrt{2}i{)}^2}$

$=\frac{1-2+1-2}{1+2}$

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{-2}{3}$ ----- ( 1 )

$\Rightarrow$ $\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=1$ --- ( 2 )

Required equation,

$x^2-(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })x+(\frac{\alpha }{\beta }\times \frac{\beta }{\alpha })=0$

$\Rightarrow$ $x^2+\frac{2}{3}x+1=0$ [ From ( 1 ) and ( 2 )]

$\therefore$ $3x^2+2x+3=0$