Question #e7e98

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Answer 1 · 2018-02-16T03:58:32

$x < 1 and x > 7$

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

Adding 3 on both sides will give us

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

Solving this equation using the quadratic formula, we get

$(8 +- sqrt(8^2 - 4(1 * 7)))/(2*1)$

This gives us
$(8 +- sqrt(64 - 28))/ 2$

Which is $(8 +- sqrt(36))/2$
Which is $(8 +- 6)/2$
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 $x < 1$