# Inequalities – Square inequality – Exercise 1700

Exercise

Solve the inequality:

$$8(1-2x)<{(2x-3)}^2$$

$$x\neq -\frac{1}{2}$$

Solution

$$8(1-2x)<{(2x-3)}^2$$

Open brackets:

$$8-16x< 4x^2-12x+9$$

Move everything to one side:

$$0<-8+16x+ 4x^2-12x+9$$

$$0<4x^2+4x+1$$

It is a square inequality. Its coefficients are

$$a=4, b=4, c=1$$

The coefficient of the square expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 4\cdot 1}}{2\cdot 4}=$$

$$=\frac{-4\pm \sqrt{0}}{8}=$$

$$=\frac{-4}{8}=-\frac{1}{2}$$

Hence, the factorizing of the equation is

$$4x^2+4x+1={(x+\frac{1}{2})}^2$$

Hence, we receive that our inequality equals

$${(x+\frac{1}{2})}^2>0$$

The left side is a squared phrase and therefore is always greater than zero except at a point where it is zero. Let’s see when it’s equal to zero:

$$x+\frac{1}{2}=0$$

$$x=-\frac{1}{2}$$

Therefore, the inequality is true for all x except this point.

In other words, inequality solution is

$$x\neq-\frac{1}{2}$$

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