Question

If $\alpha \ne \beta$ but ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$ 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. $x^2-5x+3=0$

Answer 1

Answer: (A)

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

${\alpha }^2=5\alpha -3$ ------- ( 1 )

${\beta }^2=5\beta -3$ ------- ( 2 )

Subtracting ( 2 ) from ( 1 /0,

$\Rightarrow$ ${\alpha }^2-{\beta }^2=5\alpha -3-5\beta +3$

$\Rightarrow$ $(\alpha +\beta )(\alpha -\beta )=5(\alpha -\beta )$

$\Rightarrow$ $\alpha +\beta =5$ ------- ( 3 )

Now, adding ( 1 ) and ( 2 ?),

$\Rightarrow$ ${\alpha }^2+{\beta }^2=5\alpha -3+5\beta -3$

$\Rightarrow$ ${\alpha }^2+{\beta }^2=5(\alpha +\beta )-6$

$\Rightarrow$ ${\alpha }^2+{\beta }^2=5\left(5\right)-6$

$\Rightarrow$ ${\alpha }^2+{\beta }^2=25-6$

$\therefore$ ${\alpha }^2+{\beta }^2=19$ ------ ( 4 )

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ $(5{)}^2=19+2\alpha \beta$ [ Using ( 3 ) and ( 4 ) ]

$\Rightarrow$ $25-19=2\alpha \beta$

$\Rightarrow$ $2\alpha \beta =6$

$\therefore$ $\alpha \beta =3$ ------ ( 5 )

Now,

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

$=\frac{19}{3}$ [ Using ( 4 ) and ( 5 ) ]

$\Rightarrow$ $(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })=\frac{19}{3}$ ----- ( 6 )

$\Rightarrow$ And $(\frac{\alpha }{\beta }\times \frac{\beta }{\alpha })=1$ ---- ( 7 )

The required equation,

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

$\Rightarrow$ $x^2-\frac{19}{3}x+1=0$ [ From ( 6 ) and ( 7 )]

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