Mathematico | Mathematico

The equation

\frac{x^{2}}{m + 2} = 1 - x

has a unique solution in $x .$

Find the sum of all possible values of $m$ .

Hint

Once you have found your values of $m$ , check back with the original equation to see if they make sense.

Solution

Let's get it in quadratic form

\begin{aligned} \frac{x^{2}}{m + 2} & = 1 - x \\ x^{2} & = (1 - x) (m + 2) \\ x^{2} & = m + 2 - m x - 2 x \\ x^{2} + (m + 2) x - (m + 2) & = 0 \end{aligned}

Since there is only one root, we have

\begin{aligned} b^{2} - 4 a c & = 0 \\ (m + 2)^{2} - 4 (1) (m + 2) & = 0 \\ (m + 2) (m + 2 - 4) & = 0 \\ m & = \pm 2 \end{aligned}

However, $m = - 2$ is impossible, because otherwise the denominator of $\frac{x^{2}}{m + 2}$ , which appears in the original equation, would be $0$ .

So the only answer is $m = 2$ .