Question

Let $\alpha ,\beta$ be the roots of the quadratic equation $(1-m)x^2-(2-3m)x-3m=0,m\in W$ has distinct integer roots. Then find the equation whose roots are $\alpha +5\&\beta +5$.

• A. $x^2+12x-35=0$
• B. $x^2-12x-35=0$
• C. $x^2+12x+35=0$
• D. $x^2-12x+35=0$

Answer 1

The correct option is D $x^2-12x+35=0$

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

If roots are distinct integers, then discriminant should be perfect square of an integer.
$D=b^2-4ac=(-(2-3m){)}^2-4\cdot (1-m)\cdot (-3m)$
$\Rightarrow D=4+9m^2-12m+12m-12m^2$
$\Rightarrow D=-3m^2+4$
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, $m\in W\Rightarrow m\in \{0,1\}$
If $m=0,D=-3(0{)}^2+4=4,$ which is a perfect square.
Similarly, if $m=1,D=-3(1{)}^2+4=1$ which is a perfect square.

Hence, the possible values of $m$ are $0,1$.
When $m=0$, the quadratic equation becomes $x^2-2x=0$
$\Rightarrow \alpha =0;\beta =2\cdots \left(i\right)$

When $m=1$, the equation becomes $x-3=0$ which is not a quadratic equation since coefficient of $x^2$ is $0$

Thus, we have $\alpha =0;\beta =2$

Now, we need to find the equation whose roots are $\alpha +5\&\beta +5$
$\therefore \text{roots are}0+5=5,2+5=7$
$\therefore \text{The equation is}(x-5)(x-7)=0$
$\Rightarrow x^2-12x+35=0$