Question

Let $\alpha ,\beta$ be the values of $m\in W$ for which the equation $(1-m)x^2-(2-3m)x-3m=0$ has distinct integer roots. Find the equation whose roots are $\alpha +5$ and $\beta +5$.

• A. $x^2-11x+30=0$
• B. $x^2-11x-30=0$
• C. $x^2+11x+30=0$
• D. $x^2+11x-30=0$

Answer 1

The correct option is A $x^2-11x+30=0$

Given: $(1-m)x^2-(2-3m)x-3m=0$
On comparing with standard form of quadratic equation $\text{a}{x}^2+bx+c=0$
we get, $a=(1-m),b=-(2-3m),c=-3m$

If roots are distinct integers, then discriminant should be perfect square of an integer.
$D=b^2-4ac=(-(2-3m){)}^2-4.(1-m).(-3m)$
$D=4+9m^2-12m+12m-12m^2$
$D=-3m^2+4$
Also, $D>0\Rightarrow -3m^2+4>0$
$\Rightarrow 3m^2-4<0$
$\Rightarrow (\sqrt{3}m-2)(\sqrt{3}m+2)<0$
$\Rightarrow m\in (\frac{-2}{\sqrt{3}},\frac{2}{\sqrt{3}})$

Since, $m\in W\Rightarrow m\in \{0,1\}$
So, If $m=0$, $D=-3(0{)}^2+4=4$, it is a perfect square.
$m=1$, $D=-3(1{)}^2+4=1$, it is a perfect square.

Hence, the possible values are $0,1$.
$\therefore \alpha =0;\beta =1$

We need to find the equation whose roots are $\alpha +5$ and $\beta +5$
$\therefore \text{roots are}0+5=5,1+5=6$
$\therefore \text{The equation is}(x-5)(x-6)=0$
$\Rightarrow x^2-11x+30=0$