Question

Question #e7e98

Answer 1

`x < 1 and x > 7`

Explanation:

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`