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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.

b0018877
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    Welcome to MSE community @b0018877. What is your attempt? What did you try?? – Anton Vrdoljak Oct 10 '23 at 10:28
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    Please show your own attempts. This should be a pretty short problem, perhaps two lines long. – Matti P. Oct 10 '23 at 10:33
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    Use the quadratic formula. Find the p such that the solutions differ by 6. – NicNic8 Oct 10 '23 at 13:34
  • Two numbers that differ by an integer are not necessarily integers themselves, or even rational. As you can see, searching for rational roots is useless for this particular problem, so you need to use other methods (as already suggested). – David K Oct 10 '23 at 14:18
  • One can see it this way. Suppose that $a,b$ are the two distinct roots of your equation. Then $a+b=p/2, a-b=6$ and $ab=-2$. Now try to find $p$ by using the fact that $(a+b)^2-(a-b)^2=4ab$. – ShyamalSayak Oct 20 '23 at 03:08

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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:

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

Denes Garda
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