# Difference between revisions of "1981 AHSME Problems/Problem 29"

A solution is available here. Pull up find, and put in "Since x is the principal", and you will arrive at the solution.

It's not super clear, and there's some black stuff over it, but its legible.

The solution in the above file/pdf is the following. I tried my best to match it verbatim, but I had to guess at some things. I also did not do the entire solution like this, just parts where I had to figure out what the words/math was, so this transcribed solution could be wrong and different from the solution in the aforementioned file/pdf.

Anyways:

29. (E) Since $x$ is the principal square root of some quantity, $x\geq0$. For $x\geq0$, the given equation is equivalent to $$a-\sqrt{a+x}=x^2$$ or $$a=\sqrt{a+x}+x^2.$$ The left member is a constant, the right member is an increasing function of $x$, and hence the equation has exactly one solution. We write \begin{align*} \sqrt{a+x}&=a-x^2 \\ \sqrt{a+x}+x&=(a+x)-x^2 \\ &=(\sqrt{a+x}+x)(\sqrt{a+x}-x). \end{align*}

Since $\sqrt{a+x}+x>0$, we may divide by it to obtain $$1=\sqrt{a+x}-x\quad\text{or}\quad x+1=\sqrt{a+x},$$ so $$x^2+2x+1=a+x,$$ and $$x^2+x+1-a=0.$$

Therefore $x=\frac{-1\pm\sqrt{4a-3}}{2}$, and the positive root is $x=\frac{-1+\sqrt{4a-3}}{2}$, the only solution of the original equation. Therefore, this is also the sum of the real solutions. $$\text{OR}$$

As above, we derive $a-\sqrt{a+x}=x^2$, and hence $a-x^2=\sqrt{a+x}$. Squaring both sides, we find that $$a^2-2x^2a+x^4=a+x.$$

This is a quartic equation in $x$, and therefore not easy to solve; but it is only quadratic in $a$, namely $$a^2-(2x^2+a)a+x^4-x=0.$$

Solving this by the quadratic formula, we find that \begin{align*} a&=\frac{1}{2}[2x^2+1+\sqrt{4x^4+4x^2+1-4x^4+4x}] \\ &=x^2+x+1. \end{align*} [We took the positive square root since $a>x^2$; indeed $a-x^2=\sqrt{a+x}$.]

Now we have a quadratic equation for $x$, namely $$x^2+x+1-a=0,$$ which we solve as in the previous solution.

Note: One might notice that when $a=3$, the solution of the original equation is $x=1$. This eliminates all choices except (E).

-- OliverA