# How many (real) solutions do the equation abs(x-1) = x^2 + 1 have?

## 0 1 2 3 4

May 17, 2018

3. $2$

#### Explanation:

Given:

$\left\mid x - 1 \right\mid = {x}^{2} + 1$

Trying a few small values, we can find solutions:

$\left\mid \left(\textcolor{b l u e}{0}\right) - 1 \right\mid = 1 = {\left(\textcolor{b l u e}{0}\right)}^{2} + 1$

$\left\mid \left(\textcolor{b l u e}{- 1}\right) - 1 \right\mid = 2 = {\left(\textcolor{b l u e}{- 1}\right)}^{2} + 1$

So it looks like there are $2$ solutions.

More formally and to check we have not missed any, let's try squaring the given equation to get (for real values of $x$):

${x}^{2} - 2 x + 1 = {x}^{4} + 2 {x}^{2} + 1$

Then subtracting ${x}^{2} - 2 x + 1$ from both sides we get:

$0 = {x}^{4} + {x}^{2} + 2 x$

$\textcolor{w h i t e}{0} = x \left({x}^{3} + x + 2\right)$

$\textcolor{w h i t e}{0} = x \left(x + 1\right) \left({x}^{2} - x + 2\right)$

The remaining quadratic is always positive as we can see by checking its discriminant or by completing the square:

${x}^{2} - x + 2 = {\left(x - \frac{1}{2}\right)}^{2} + \frac{7}{4}$

So it has no factors with real coefficients.