Question

$x^2-\frac{k}{2}x=2p$
In the quadratic equation above, k and p are constants. What are the solutions for x?

• A. $x=\frac{k}{4}\pm \frac{\sqrt{k^2+2p}}{4}$
• B. $x=\frac{k}{4}\pm \frac{\sqrt{k^2+32p}}{4}$
• C. $x=\frac{k}{2}\pm \frac{\sqrt{k^2+2p}}{2}$
• D. $x=\frac{k}{2}\pm \frac{\sqrt{k^2+32p}}{4}$

Answer 1

Answer: (B)

$x=\frac{k}{4}\pm \frac{\sqrt{k^2+32p}}{4}$

We know that for a quadratic equation of the form $a{x}^2+bx+c=0$,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ [Quadratic formula]

The given equation is $x^2-\frac{k}{2}x=2p$

We can rewrite the equation as $x^2-\frac{k}{2}x-2p=0$

$\therefore x=\frac{\frac{k}{2}\pm \sqrt{{\left(\frac{-k}{2}\right)}^2-4\times 1\times (-2p)}}{2\times 1}$ [Quadratic formula]

$\Rightarrow \frac{\frac{k}{2}\pm \sqrt{\frac{k^2}{4}+8p}}{2}$

$\Rightarrow \frac{k\pm \sqrt{k^2+32p}}{4}$

$\Rightarrow \frac{k}{4}\pm \frac{\sqrt{k^2+32p}}{4}$

Hence, option B is correct.