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

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==Problem==
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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>
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is equal to
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<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>
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==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 $a > 1$, 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

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

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