1981 AHSME Problems/Problem 29

Problem

If $a > 1$ , then the sum of the real solutions of

$\sqrt{a - \sqrt{a + x}} = x$

is equal to

$\textbf{(A)}\ \sqrt{a} - 1\qquad \textbf{(B)}\ \dfrac{\sqrt{a}- 1}{2}\qquad \textbf{(C)}\ \sqrt{a - 1}\qquad \textbf{(D)}\ \dfrac{\sqrt{a - 1}}{2}\qquad \textbf{(E)}\ \dfrac{\sqrt{4a- 3} - 1}{2}$

Solution 1

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.

-OliverA

Solution 2

Recall that the square root of a real number cannot be negative, so $x \geq 0$ . Square both sides of the equation to get $a - \sqrt{a + x} = x^2$ , which is equivalent to $a - x^2 = \sqrt{a+x}$ .

Once again, the square root of a real number cannot be negative, so $a - x^2 \geq 0$ , that is $x^2 \geq a$ . So $-\sqrt{a} \leq x \leq \sqrt{a}$ , and since $x \geq 0$ as previously determined, $0 \leq x \leq \sqrt{a}$ .

Squaring both sides again yields $a^2 - 2ax^2 + x^4 = a + x$ . Collecting all terms on the yields $a^2 - 2ax^2 - a - x^4 - x =$ $a^2 - (2x^2 + 1) a + (x^4 - x) = 0$ .

Using the quadratic formula, $a = \frac{2x^2+1 \pm \sqrt{(2x^2+1)^2 - 4(x^4 -x)}}{2} =$ $\frac{2x^2 + 1 \pm \sqrt{4x^2 + 4x + 1})}{2} = \frac{2x^2 + 1 \pm (2x + 1)}{2}$ , so either $a = x^2 + x + 1$ or $a = x^2 - x$ .

If $a = x^2 + x + 1$ , then $x^2 + x + (1-a) = 0$ , and by the quadratic formula, $x = \frac{-1 \pm \sqrt{1 - 4(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}$ . Since $\frac{-1-\sqrt{4a-3}}{2}$ is negative, it cannot a valid solution. However, $\frac{-1+\sqrt{4a-3}}{2} \geq \frac{-1+1}{2} = 0$ , and $\frac{-1+\sqrt{4a-3}}{2} < {-1 + 2\sqrt{a}}{2} = -\frac{1}{2} + \sqrt{a} < \sqrt{a}$ , so $\frac{-1+\sqrt{4a-3}}{2}$ is a valid solution.

If $a = x^2 - x$ , then $x^2 - x - a = 0$ , and by the quadratic formula, $x = \frac{1 \pm \sqrt{1+4a}}{2}$ . But $\frac{1-\sqrt{1+4a}}{2} < {1-1}{2} = 0$ , and $\frac{1+\sqrt{1+4a}{2}} > \frac{\sqrt{1+4a}}{2} > \frac{\sqrt{4a}}{2} = \sqrt{a}$ , so neither of these are valid solutions.

Therefore, $x = \frac{-1+\sqrt{4a-3}}{2}$ is the only solution to the original equation, so the sum of all solutions is $\boxed{(\textbf{E})\ \frac{-1+\sqrt{4a-3}}{2}}$ .

-OliverA, edited by j314andrews. Similar to the alternate solution from The Contest Problem Book IV.

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

-- OliverA

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 29

Contents

Problem

Solution 1

Solution 2

See also