Question

If $\alpha ,\beta$ are the roots of the equation $x^2-ax+b=0$ and roots of the eqaution $b{x}^2-4x+4=0$ are $\alpha +\frac{{\beta }^2}{\alpha }$ , $\beta +\frac{{\alpha }^2}{\beta }$ then$a+b$ can be -

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B), (D)

The correct options are
B $1$
D $3$

$x^2+ax+b=0$
$\alpha \beta =b,\alpha +\beta =a$
Roots of the equation $b{x}^2-4x+4=0$ are $\alpha +\frac{{\beta }^2}{\alpha }$ , $\beta +\frac{{\alpha }^2}{\beta }$
$\alpha +\frac{{\beta }^2}{\alpha }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha }=\frac{a^2-2b}{\alpha }\beta +\frac{{\alpha }^2}{\beta }=\frac{a^2-2b}{\beta }$
So, the equation having these two roots will be
$x^2+\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$
$\Rightarrow x^2-\frac{a(a^2-2b)}{b}x+\frac{(a^2-2b{)}^2}{b}=0b{x}^2-a(a^2-2b)x+(a^2-2b{)}^2=\mathrm{0............}\left(i\right)b{x}^2-4x+4=\mathrm{0................}\left(ii\right)$
Comparing $\left(i\right)$ and $\left(ii\right)$ we have
$a^2-2b=\pm 2\&a(a^2-2b)=4a=2\&b=1a=-2\&b=3$