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

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+ | ==Problem== | ||

+ | If <math> a > 1</math>, then the sum of the real solutions of | ||

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+ | <math> \sqrt{a - \sqrt{a + x}} = x</math> | ||

+ | |||

+ | is equal to | ||

+ | |||

+ | <math> \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}</math> | ||

+ | |||

+ | |||

+ | ==Solution== | ||

A solution is available [https://files.eric.ed.gov/fulltext/ED239856.pdf here]. Pull up find, and put in "Since x is the principal", and you will arrive at the solution. | A solution is available [https://files.eric.ed.gov/fulltext/ED239856.pdf here]. Pull up find, and put in "Since x is the principal", and you will arrive at the solution. | ||

## Latest revision as of 12:10, 18 July 2020

## Problem

If , then the sum of the real solutions of

is equal to

## Solution

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 is the principal square root of some quantity, . For , the given equation is equivalent to or The left member is a constant, the right member is an increasing function of , and hence the equation has exactly one solution. We write

Since , we may divide by it to obtain so and

Therefore , and the positive root is , the only solution of the original equation. Therefore, this is also the sum of the real solutions.

As above, we derive , and hence . Squaring both sides, we find that

This is a quartic equation in , and therefore not easy to solve; but it is only quadratic in , namely

Solving this by the quadratic formula, we find that [We took the positive square root since ; indeed .]

Now we have a quadratic equation for , namely which we solve as in the previous solution.

*Note*: One might notice that when , the solution of the original equation is . This eliminates all choices except (E).

-- OliverA