Solve quadratic with unknown constant in question

Question

The quadratic equation $2x^2-px-4=0$ where $p$ is a positive constant, has 2 solutions that differ by 6. What is the value of $p$?

I attempted to factorise to get (2x-2)(x+2)=0 and (2x-4)(x+1)=0 but the solutions don’t differ correctly.

Answer 1

Using the quadratic formula this is really easy to solve.

We know that one solution will be $\frac{p + \sqrt{p^2+32}}{4}$ and the other one will be $\frac{p - \sqrt{p^2+32}}{4}$. We know the first one is going to be bigger than the second one because it adds the discriminant instead of subtracting it. Using this information and the fact that the difference between the solutions must be $6$, we can get the following equation:

$$ \frac{p + \sqrt{p^2+32}}{4} - \frac{p - \sqrt{p^2+32}}{4} = 6 $$

All that's left to do is solve for $p$:

$\frac{p + \sqrt{p^2+32}}{4} - \frac{p - \sqrt{p^2+32}}{4} = 6 \\ p + \sqrt{p^2+32} - p + \sqrt{p^2+32} = 24 \\ \sqrt{p^2+32} + \sqrt{p^2+32} = 24 \\ 2\sqrt{p^2+32} = 24 \\ \sqrt{p^2+32} = 12 \\ p^2 + 32 = 144 \\ p^2 = 112 \\ p = \pm 4 \sqrt{7}$

We can test for any extraneous solutions, but after plugging both values into the original quadratic and seeing that they both work, we can conclude that:

$$ \boxed{p= \pm 4\sqrt{7}} $$