Exercise
Solve the inequality:
8(1-2x)<{(2x-3)}^2
Final Answer
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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