An Equation in Radicals

Here's an engaging and a rather nonstandard problem from the 1981 American High School Mathematics Examination (#29)

Solution

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Copyright © 1996-2012 Alexander Bogomolny

If \(a\ \ge 1\), then the sum of the real solutions of \(\sqrt{a-\sqrt{a+x}}=x\) is equal to

(A) \(\sqrt{a}-1\)  (B) \(\frac{\sqrt{a}-1}{2}\)  (C)\(\sqrt{a-1}\)  (D) \(\frac{\sqrt{a-1}}{2}\)  (E) \(\frac{\sqrt{4a-3}-1}{2}\)

We'll give three solutions to the problem. Two of the solutions are from the The Contest Book.

First observe that squaring twice leads to a quartic (of degree 4) equation:

(1)

\(a^{2}-2x^{2}a+x^{4}=a+x\).

Such an equation may have 0, 2, or 4 real roots - an even number of roots anyway. By Viète's formulas the sum of all the roots of that polynomial is 0, but who knows whether all (or even any) of them are real. Checking whether all are real appears as difficult as the problem itself. And, since the AMCs are run in a rigid time frame, the problem may need a not so straightforward approach.

One indication that there may be no need to tackle a quartic equation is the following observation. Square the equation once and rewrite the result as

\(\sqrt{a + x} + x^{2} = a.\)

On the left, there is a strictly increasing function of \(x\); on the right a constant. It follows that, if the equation has a real solution at all, that solution is going to be unique. In that case, that number will answer the problem. (If there is a solution, the problem will serve a non-trivial example of the appearance of spurious roots due to squaring.)

Just in case we may need this, note that the form of the original equation (1) imposes several constraints on \(x\). Since \(x\) is required to be a square root of a real expression, it is non-negative. Also, the expression under the outer square root can't be negative either:

\(0 \le x, 0 \le a+x \le a^{2}.\)

Solution 1

Put the equation into the form

\(\sqrt{a + x} = a-x^{2},\)

add \(x\) to both sides and factor (this is an insightful and crucial step):

\(\sqrt{a + x} + x = a + x -x^{2} = (\sqrt{a + x} + x)(\sqrt{a + x} - x).\)

Since \(x \ge 0\), and \(a\ge 1\), \(\sqrt{a + x} + x \gt 0\), we can divide by the left-hand side \(\sqrt{a + x} + x\) to obtain

\(1 = \sqrt{a + x} - x,\)

or, equivalently,

\(\sqrt{a + x} = x + 1.\)

Now, squaring leads to a quadratic equation:

\(a + x = x^{2} + 2x + 1\), or

or,

(2)

\(x^{2}+x+(1-a)=0.\)

The quadratic formula then gives the roots

\(x_{1,2}=\frac{-1\pm\sqrt{1-4(1-a)}}{2}\),

in which only one is not negative:

\(x=\frac{-1+\sqrt{4a-3}}{2}\),

and this gives the answer to our problem.

Solution 2

Observe that, although (1) is a quartic equation in \(x\), it is quadratic in \(a\):

\(a^{2}-a(2x^{2}+1)+(x^{4}-x)=0\).

Thus, the quadratic formula yields a relationship between \(a\) and \(x\)

\(a_{1,2}=\frac{(2x^{2}+1)\pm\sqrt{(2x^{2}+1)^{2}-4(x^{4}-x)}}{2}=\frac{(2x^{2}+1)\pm\sqrt{4x^{2}+4x+1}}{2}=\frac{(2x^{2}+1)\pm(2x+1)}{2}.\)

Since, \(a\ge 1\), only one relationships legitimate:

\(a = x^{2}+x+1.\)

This leads us back to the quadratic (in \(x\)) equation (2), with the same conclusions.

Solution 3

This is - in a sense - a rephrase of Solution 1 that leads to the requisite factoring in a more (in my view) transparent way.

Introduce (provisionally) \(y=\sqrt{a+x}\) and consider a system of two equations

\( \begin{align} y&=\sqrt{a+x}\\ x&=\sqrt{a-y} \end{align} \).

Squaring gives

\( \begin{align} y^{2}&=a+x \\ x^{2}&=a-y \end{align} \).

Subtract the second equation from the first:

\(y^{2}-x^{2}=(y+x)(y-x)=x+y.\)

Dividing by \(y+x\)

\(y-x=1\).

Substituting back \(y=\sqrt{a+x}\) returns us to equation (2).

Reference

1. R. A. Artino, A. M. Gaglione, N. Shell, The Contest Problem Book IV (AHSME 1973-1082), MAA, December 1982

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Copyright © 1996-2012 Alexander Bogomolny