# Solving a quadratic formula with positive discriminant yields only one correct solution.

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.)

• Did you try substituting $p=9$ in your second written equation? And in your third one? – Arnaud D. Nov 9 '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 – Vasya Nov 9 '18 at 17:21
• Vasya, wouldn't leaving $p < 15$ allow for an imaginary solution? – Natalie Webb Nov 9 '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 9 '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 9 '18 at 18:41

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, 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 9 '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 9 '18 at 20:30