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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.)

Natalie Webb
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  • Did you try substituting $p=9$ in your second written equation? And in your third one? – Arnaud D. Nov 09 '18 at 17:20
  • To get rid of the square root you square both parts. It's OK to do this only when both part have the same sign, otherwise you can get extra solutions (i.e. $(-2)^2=2^2$). You have $-2\sqrt{p}=15 - p$ which should give you $p \ge 15$ as a condition for possible roots because LHS is non-positive – Vasili Nov 09 '18 at 17:21
  • Vasya, wouldn't leaving $p < 15$ allow for an imaginary solution? – Natalie Webb Nov 09 '18 at 18:09
  • Because requiring $p \geq 15$ would imply the assumption that $-2\sqrt{p}$ is non-positive, not considering that $\sqrt{p}$ may itself be negative. – Natalie Webb Nov 09 '18 at 18:22
  • The quadratic formula is irrelevant. It's important to realize that squaring both sides of an equation can introduce extraneous solutions. Indeed, this is a crucial part of this topic (solving radical equations). See this google search for more information. – Dave L. Renfro Nov 09 '18 at 18:41

1 Answers1

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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 $(\sqrt p-5)(\sqrt p+3)=0$.

Mike
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  • Mike, what you're saying is that factoring to $(\sqrt{p} - 5)(\sqrt{p} + 3)$ results in $\sqrt{p} = -3, 5$, but $\sqrt{p}$ as a function must return a positive number, meaning that $\sqrt{p} = -3$ can't be a solution? – Natalie Webb Nov 09 '18 at 18:51
  • @NatalieWebb That's a fair way of putting it. The basic idea is you added another root when you squared both sides. If $(15-p)^2-4p=0$, then $(15-p-2\sqrt p)(15-p+2\sqrt p)=0$ – Mike Nov 09 '18 at 20:30