Question

The number of values of $c\in N$ for which the the quadratic equation $x^2-13x+4c=0$ has distinct integer roots are:

• A. $3$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

Given equation is $x^2-13x+6c=0$

On comparing with the standard quadratic equation $a{x}^2+bx+c=0,a\ne 0$
we get, $a=1;b=-13;c=6c$
Using discriminant method,
$x=\frac{-b\pm \sqrt{D}}{2a}$
If roots are distinct integers, then discriminant should be perfect square of an integer.
Now, $D=b^2-4ac=(-13{)}^2-\mathrm{4.1.6}c$
$D=169-24c$
Also, $D>0\Rightarrow 169-24c>0\Rightarrow c<\frac{169}{24}$
Since, $c\in N\Rightarrow c\text{must be an element of set}\{1,2,3,4,5,6,7\}$
So, If $c=1$, $D=169-24\left(1\right)=145$, which is not a perfect square.
$c=2$, $D=169-24\left(2\right)=121$, it is a perfect square .
$c=3$, $D=169-24\left(3\right)=97$, which is not a perfect square.
$c=4$, $D=169-24\left(4\right)=73$,which is not a perfect square.
$c=5$, $D=169-24\left(5\right)=49$ it is a perfect square .
$c=6$, $D=169-24\left(6\right)=25$, it is a perfect square .
$c=7$, $D=169-24\left(7\right)=1$, it is a perfect square .

Hence, the possible values are $2,5,6,7$. $\therefore$ The number of possible values of $c$ are $4$.