Question

If $\alpha ,\beta$ are roots of the equation $6x^2-x-2=0$, then the equation whose roots are ${\alpha }^2+2,{\beta }^2+2$, is

• A. $36x^2-169x+198=0$
• B. $36x^2+169x+198=0$
• C. $36x^2-169x-198=0$
• D. $x^2-169x+198=0$

Answer 1

Answer: (A)

Since, $\alpha ,\beta$ are roots of $6x^2-x-2=0$
$\Rightarrow \alpha -1/2,\beta =\frac{2}{3}$
Now, ${\alpha }^2+2=9/4$
${\beta }^2+2=\frac{22}{9}$
$S={\alpha }^2+{\beta }^2+4=\frac{169}{36}$
$P=({\alpha }^2+2)({\beta }^2+2)=\frac{198}{36}$
$\therefore$ Required equation is $x^2-\frac{169}{36}x+\frac{198}{36}=0$
or $36x^2-169x+198=0$
Alternative Solution : Given equation is $6x^2-x-2=0$
$S=\alpha +\beta =\frac{1}{6},\alpha \cdot \beta =-\frac{1}{3}$
Now, equation whose roots are ${\alpha }^2+2,{\beta }^2+2$, is
$x^2-({\alpha }^2+{\beta }^2+4)x+({\alpha }^2+2)({\beta }^2+2)=0$
or $x^2-[(\alpha -\beta {)}^2-2\alpha \beta +4]x+[{\alpha }^2{\beta }^2+2({\alpha }^2+{\beta }^2)+4]=0$
or $x^2-\frac{169}{36}+\frac{198}{36}=0,36x^2-169x+198=0$
Short Cut Method :
We have to determine the equation whose roots are
${\alpha }^2+2,{\beta }^2+2$
let $t={\alpha }^2+2$
${\alpha }^2=t-2\dots (i)$
Now, from given equation
$6x^2-2=x$
$\therefore$ squaring both sides, we get
$[6(x^2)-2{]}^2=x^2$
$[6(t-2)-2{]}^2=t-2$ (Using (i))
or $(6t-14{)}^2=t-2$
or $36t^2-169t+198=0$
or $36x^2-169x+198=0$