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Inequalities – Square inequality – Exercise 1700

Exercise 

Solve the inequality:

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

Final Answer


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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