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Q.
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
Complex Numbers and Quadratic Equations
Solution:
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 $36x^{2}-169x+198=0$