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

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Using the quadratic formula, <math>a = \frac{2x^2+1 \pm \sqrt{(2x^2+1)^2 - 4(x^4 -x)}}{2} = </math> <math>\frac{2x^2 + 1 \pm \sqrt{4x^2 + 4x + 1})}{2} = \frac{2x^2 + 1 \pm (2x + 1)}{2}</math>, so either <math>a = x^2 + x + 1</math> or <math>a = x^2 - x</math>.
 
Using the quadratic formula, <math>a = \frac{2x^2+1 \pm \sqrt{(2x^2+1)^2 - 4(x^4 -x)}}{2} = </math> <math>\frac{2x^2 + 1 \pm \sqrt{4x^2 + 4x + 1})}{2} = \frac{2x^2 + 1 \pm (2x + 1)}{2}</math>, so either <math>a = x^2 + x + 1</math> or <math>a = x^2 - x</math>.
  
−
If <math>a = x^2 + x + 1</math>, then <math>x^2 + x + (1-a) = 0</math>, and by the quadratic formula, <math>x = \frac{-1 \pm \sqrt{1 - 4(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}</math>.  Since <math>\frac{-1-\sqrt{4a-3}}{2}</math> is negative, it cannot a valid solution.  However, <math>\frac{-1+\sqrt{4a-3}}{2} \geq \frac{-1+1}{2} = 0</math>, and <math>\frac{-1+\sqrt{4a-3}}{2} < {-1 + 2\sqrt{a}}{2} = -\frac{1}{2} + \sqrt{a} < \sqrt{a}</math>, so <math>\frac{-1+\sqrt{4a-3}}{2}</math> is a valid solution.
+
If <math>a = x^2 + x + 1</math>, then <math>x^2 + x + (1-a) = 0</math>, and by the quadratic formula, <math>x = \frac{-1 \pm \sqrt{1 - 4(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}</math>.  Since <math>\frac{-1-\sqrt{4a-3}}{2}</math> is negative, it cannot a valid solution.  However, <math>\frac{-1+\sqrt{4a-3}}{2} \geq \frac{-1+1}{2} = 0</math>, and <math>\frac{-1+\sqrt{4a-3}}{2} < \frac{-1 + 2\sqrt{a}}{2} = -\frac{1}{2} + \sqrt{a} < \sqrt{a}</math>, so <math>\frac{-1+\sqrt{4a-3}}{2}</math> is a valid solution.
  
−
If <math>a = x^2 - x</math>, then <math>x^2 - x - a = 0</math>, and by the quadratic formula, <math>x = \frac{1 \pm \sqrt{1+4a}}{2}</math>.  But <math>\frac{1-\sqrt{1+4a}}{2} < {1-1}{2} = 0</math>, and <math>\frac{1+\sqrt{1+4a}{2}} > \frac{\sqrt{1+4a}}{2} > \frac{\sqrt{4a}}{2} = \sqrt{a}</math>, so neither of these are valid solutions.
+
If <math>a = x^2 - x</math>, then <math>x^2 - x - a = 0</math>, and by the quadratic formula, <math>x = \frac{1 \pm \sqrt{1+4a}}{2}</math>.  But <math>\frac{1-\sqrt{1+4a}}{2} < \frac{1-1}{2} = 0</math>, and <math>\frac{1+\sqrt{1+4a}}{2} > \frac{\sqrt{1+4a}}{2} > \frac{\sqrt{4a}}{2} = \sqrt{a}</math>, so neither of these are valid solutions.
  
 
Therefore, <math>x = \frac{-1+\sqrt{4a-3}}{2}</math> is the only solution to the original equation, so the sum of all solutions is <math>\boxed{(\textbf{E})\ \frac{-1+\sqrt{4a-3}}{2}}</math>.
 
Therefore, <math>x = \frac{-1+\sqrt{4a-3}}{2}</math> is the only solution to the original equation, so the sum of all solutions is <math>\boxed{(\textbf{E})\ \frac{-1+\sqrt{4a-3}}{2}}</math>.

Revision as of 14:53, 19 August 2025

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 $\frac{-1+\sqrt{4a-3}}{2} < \frac{-1 + 2\sqrt{a}}{2} = -\frac{1}{2} + \sqrt{a} < \sqrt{a}$, 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 $\frac{1-\sqrt{1+4a}}{2} < \frac{1-1}{2} = 0$, 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

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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