Which of the following condition imply that roots of the equation $\left(\frac{1}{4}\right) x^2+b x+c=0$ are…
Which of the following condition imply that roots of the equation $\left(\frac{1}{4}\right) x^2+b x+c=0$ are integers?
$b^2-c>0$
$b$ and $c$ are even integers
$b^2-c$ is the square of an integer and $b$ is an integer
$b$ and $c$ are integers
Solution
We have,
$\frac{1}{4} x^2+b x+c=0$
By using quadratic formula
$x=\frac{-b \pm \sqrt{b^2-4 \times \frac{1}{4} \times c}}{2 \times \frac{1}{4}}$
$x=\frac{-b \pm \sqrt{b^2-c}}{\frac{1}{2}}$
The roots are integer iff $b$ is an integer and $b^2-c$ is perfect square.