Question #e7e98

1 Answer
Feb 16, 2018

Answer:

#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#