# Inequalities – Square inequality – Exercise 5700

Exercise

Solve the square inequality:

$$4x^2-12x\geq -10$$

All x

Solution

$$4x^2-12x\geq -10$$

Move everything to one side:

$$4x^2-12x+10\geq 0$$

It is a square inequality. its coefficients are

$$a=4, b=-12, c=10$$

The coefficient of the squared 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

$$x_{1,2}=\frac{12\pm \sqrt{{(-12)}^2-4\cdot 4\cdot 10}}{2\cdot 4}=$$

$$=\frac{12\pm \sqrt{-16}}{8}$$

We got a negative number inside the root, so there is no real solution to the quadratic equation, i.e. its graph does not pass through the x-axis. Because it is “smiling”, it is always above the x-axis. Hence, the solution of the inequality is all x.

The graph of the equation:

$$y=4x^2-12x+10$$

looks like this: You can see that the graph is indeed above the x-axis for all x.

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