# Solve the following ? Please

## $| {x}^{2} - 4 x + 3 | = x + 1$

Jun 17, 2018

$x = \frac{5 \pm \sqrt{17}}{2}$

#### Explanation:

If $| {x}^{2} - 4 x + 3 | = x + 1$, we have $x \ge - 1$ as $| {x}^{2} - 4 x + 3 | \ge 0$. This gives us the domain of $x$.

Now we have two possibilities

${x}^{2} - 4 x + 3 = x + 1$ or ${x}^{2} - 5 x + 2 = 0$ i.e.

$x = \frac{5 \pm \sqrt{{5}^{2} - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}$

Observe that both are in domain.

If ${x}^{2} - 4 x + 3 = - x - 1$ or ${x}^{2} - 3 x + 4 = 0$. But here discriminant is negative 3^2-4×1×4=-7. Hence we do not have any solution.