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

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− | x | + | 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. |

+ | |||

+ | 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 <math>x</math> is the principal square root of some quantity, <math>x\geq0</math>. For <math>x\geq0</math>, the given equation is equivalent to <cmath>a-\sqrt{a+x}=x^2</cmath> or <cmath>a=\sqrt{a+x}+x^2.</cmath> The left member is a constant, the right member is an increasing function of <math>x</math>, and hence the equation has exactly one solution. We write | ||

+ | <cmath>\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*}</cmath> | ||

+ | |||

+ | Since <math>\sqrt{a+x}+x>0</math>, we may divide by it to obtain <cmath>1=\sqrt{a+x}-x\quad\text{or}\quad x+1=\sqrt{a+x},</cmath> | ||

+ | so <cmath>x^2+2x+1=a+x,</cmath> and <cmath>x^2+x+1-a=0.</cmath> | ||

+ | |||

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

+ | <cmath>\text{OR}</cmath> | ||

+ | |||

+ | As above, we derive <math>a-\sqrt{a+x}=x^2</math>, and hence <math>a-x^2=\sqrt{a+x}</math>. Squaring both sides, we find that <cmath>a^2-2x^2a+x^4=a+x.</cmath> | ||

+ | |||

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

+ | |||

+ | Solving this by the quadratic formula, we find that | ||

+ | <cmath>\begin{align*} | ||

+ | a&=\frac{1}{2}[2x^2+1+\sqrt{4x^4+4x^2+1-4x^4+4x}] \\ | ||

+ | &=x^2+x+1. | ||

+ | \end{align*}</cmath> | ||

+ | [We took the positive square root since <math>a>x^2</math>; indeed <math>a-x^2=\sqrt{a+x}</math>.] | ||

+ | |||

+ | Now we have a quadratic equation for <math>x</math>, namely <cmath>x^2+x+1-a=0,</cmath> | ||

+ | which we solve as in the previous solution. | ||

+ | |||

+ | ''Note'': One might notice that when <math>a=3</math>, the solution of the original equation is <math>x=1</math>. This eliminates all choices except (E). | ||

+ | |||

+ | -- OliverA |

## Revision as of 12:07, 18 July 2020

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