Question

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

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

Answer 1

Answer: (B)

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

$x^2-2x+3=0$
$x^2-(\frac{2}{\beta }+\frac{\beta }{\propto })x+120$
$x^2-\left[\frac{(2+\beta 1^2-2\propto \beta }{\propto \beta }\right]+1=0$
$\propto +\beta =2$
$\propto \beta =3$
$x^2-\left(\frac{(2^2-6)}{3}\right)x+1=0$
$3x^2+2x+3=0$