By considering the equation \(x^2+x-a=0\,\),
show that the equation
\(x={(a-x)\vphantom M}^{\frac12}\) has one real solution when \(a\ge0\) and no
real solutions when \(a<0\,\).
Find the number of distinct real solutions of the equation
\[
x={\big((1+a)x-a\big)}^{\!\frac13}
\]
in the cases that arise according to the value of \(a\).
Find the number of distinct real solutions of the equation
\[
x={(b+x)\vphantom M}^{\frac12}
\]
in the cases that arise according to the value of \(b\,\).
Solution:
\(\,\)
\begin{align*}
&& x &= (a-x)^{\frac12} \\
\Rightarrow && x^2 &= a - x \\
\Rightarrow && 0 &= x^2 + x - a
\end{align*}
This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise.
\begin{align*}
&& x & = \left ( (1+a)x - a \right)^{\frac13} \\
\Leftrightarrow && x^ 3 &= (1+a)x - a \\
\Leftrightarrow && 0 &= x^3- (1+a)x + a \\
\Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\
\end{align*}
Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore:
One solution if \(a < -\frac14\)
Two solutions if \(a = -\frac14, 2\)
Three solutions if \(a > -\frac14, a \neq 2\)
\(\,\) \begin{align*}
&& x &= (b+x)^{\frac12} \\
\Rightarrow && x^2 &= b + x \\
\Rightarrow && 0 &= x^2 - x - b
\end{align*}
This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\).
Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)