Quant interview problem

The following is a sample problem from a quant interview presented in Isichenko [2021] (see page 62).

Try to keep your thinking flexible while attempting to solve the equation. The solution does not require anything more than what you learned about the quadratic equation in high school.

Exercise

Solve for x:

\[\sqrt{a-\sqrt{a+x}}=x.\]

Solution

Begin by getting rid of the square-roots:

\[\begin{gather*} a-\sqrt{a+x} = x^{2} \\ a - x^2 = \sqrt{a+x} \\ a^2 - 2ax^2 + x^4 = a + x. \end{gather*}\]

At this point it may not be so obvious how to proceed. But if we try to think flexibly, we might notice that if we treat \(a\) as the variable and \(x\) as the parameter (rather than the more natural other way around), we can reorganize the equation as

\[a^2 - \left(1 + 2 x^2\right) a + (x^4 - x) = 0\]

so it fits the standard form of a quadratic equation in \(a\)

\[\alpha a^2 + \beta a + \gamma=0,\]

with

\[\alpha=1, \quad \beta=-\left(1 + 2 x^{2}\right), \quad \gamma=x^{4}-x.\]

Solving for \(a\) with the roots of a quadratic equation,

\[a=\frac{\left(1 + 2 x^{2}\right) \pm \sqrt{\left(1 + 2 x^{2}\right)^{2}-4\left(x^{4}-x\right)}}{2}\]

The part under the square-root is

\[\begin{align*} \left(1 + 2 x^{2}\right)^{2}-4\left(x^{4}-x\right) &= 1 + 4x^2 + 4x^4 - 4x^4 + 4x \\ &= (2x+1)^2 \end{align*}\]

Therefore,

\[\begin{equation*} a = \frac{2x^2 + 1 \pm (2x+1)}{2} = \left\{\begin{matrix} x^{2}+x+1 \\ x^{2}-x \end{matrix}\right., \end{equation*}\]

so either

\[x^{2}+x+1-a=0\]

or

\[x^{2}-x-a=0.\]

The solutions to these two equations are, respectively,

\[x = \frac{-1 \pm \sqrt{1-4(1-a)}}{2} = \frac{-1 \pm \sqrt{4a-3}}{2}\]

and

\[x=\frac{1 \pm \sqrt{1+4a}}{2}.\]

For real solutions, \(x\) must be non-negative, so the first equation implies

\[x = \frac{\sqrt{4a-3} - 1}{2}, \quad a\geq 1.\]

Note that \(a=0\), \(x=0\) is also a solution. \(\blacksquare\)

As a quick verification, choose \(a=7\). Then

\[x = \frac{\sqrt{4a-3} - 1}{2} = 2\]

and it is easy to verify that

\[\sqrt{7 - \sqrt{7 + 2}} = 2.\]

Sympy can also solve problems like this:

Show code cell content

Hide code cell content

from sympy.solvers import solve
from sympy import Symbol, sqrt

a = Symbol('a')
x = Symbol('x')

solve(sqrt(a - sqrt(a+x)) - x, x)

[1/2 - sqrt(4*a + 1)/2,
 -sqrt(4*a - 3)/2 - 1/2,
 sqrt(4*a - 3)/2 - 1/2,
 sqrt(4*a + 1)/2 + 1/2]