# Question #e7e98

Feb 16, 2018

$x < 1 \mathmr{and} x > 7$

#### Explanation:

Given ${x}^{2} - 8 x + 4 > - 3$

Adding 3 on both sides will give us

${x}^{2} - 8 x + 7 > 0$

Solving this equation using the quadratic formula, we get

$\frac{8 \pm \sqrt{{8}^{2} - 4 \left(1 \cdot 7\right)}}{2 \cdot 1}$

This gives us
$\frac{8 \pm \sqrt{64 - 28}}{2}$

Which is $\frac{8 \pm \sqrt{36}}{2}$
Which is $\frac{8 \pm 6}{2}$
Which gives 2 solutions:

$x = 7$ and $x = 1$

Now check whether the inequality is valid for values greater or lesser than the ones we have obtained from the quadratic equation.

Substituting $x = 8$ we can see that the original inequality given in the question is true, so x must be greater than 7

$x > 7$

And taking a value less than 1, suppose 0, also shows that the inequality in the question is true.

So $x < 1$