I'm a math tutor at a small university. One of my students asked me about the problem, $$p – 2\sqrt{p} = 15$$ Solving this, we found, in sequence, $$-2\sqrt{p}=15 – p$$ $$4p = p^2 – 30 p + 225$$ $$p^2 – 34 p + 225 = 0$$ then, using quadratic formula, $$p = \frac{-(-34) \pm \sqrt{34^2 – 4(1)(225)}}{2(1)}$$ $$p = \frac{34 \pm \sqrt{256}}{2}$$
Note the positive discriminant suggesting two solutions. Solving resulted in $$p = \frac{34 \pm 16}{2} = 25, 9$$
Checking the solution, we have $$25 – 2\sqrt{25} = 15$$ $$15 = 15$$ and then $$9 – 2\sqrt{9} = 15$$ $$3 = 15$$
And I can't for the life of me figure out why 9 doesn't solve the initial equation, despite being a solution given by the quadratic formula. Looking at the graph of $y = p^2 – 34p + 225$ shows that 9 and 25 should be solutions, but the graph of $y = p – 2\sqrt{p} – 15$ has only one solution at $p = 25$, and is not remotely equivalent to the first graph. What changed? In addition, the first graph suggests to me that $p – 2\sqrt{p} = 15$ might have an additional, but imaginary, solution; but I have no idea how I might find it.
I googled around for about thirty minutes, and didn't find anyone asking the same question. This question about mechanics appears similar but doesn't yield a satisfactory answer as to why one solution doesn't work. (Typical physicists, am I right my friends? Heehee.)
Best Answer
Note that $\sqrt p$ means the positive square root of $p$. The negative square root of $9$ would be $-3$, which would have fulfilled the equation. The original equation also could have been factored as $(\sqrt p-5)(\sqrt p+3)=0$.