# Inequalities – Quadratic equation with a parameter – Exercise 5723

Exercise

Given the equation:

$$3(2m+x)=2(mx-1)$$

For which values of parameter m, the line is below the x-axis.

$$-\frac{1}{2}

Solution

Given the equation:

$$3(2m+x)=2(mx-1)$$

Open brackets:

$$6m+3x=2mx-2$$

Isolate x on one side:

$$x=\frac{6m+2}{2m-3}$$

We want to find out when the expression we got is less than zero, meaning

$$\frac{6m+2}{2m-3}<0$$

Solving inequality of a fraction is equivalent to solving an inequality of a multiplication, so instead of solving the inequality we got, we will solve the inequality:

$$(6m+2)(2m-3)<0$$

Open brackets:

$$12m^2-18m+4m-6<0$$

$$12m^2-14m-6<0$$

Reduce by 2:

$$6m^2-7m-3<0$$

It is a square inequality. its coefficients are:

$$a=6, b=-7, c=-3$$

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 below 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

$$x_{1,2}=\frac{7\pm \sqrt{{(-7)}^2-4\cdot 6\cdot (-3)}}{2\cdot 6}=$$

$$=\frac{7\pm \sqrt{121}}{12}=$$

$$=\frac{7\pm 11}{12}$$

Hence, we get the solutions:

$$x_1=\frac{7+ 11}{12}=\frac{18}{12}=\frac{3}{2}$$

$$x_2=\frac{7- 11}{12}=\frac{-4}{12}=-\frac{1}{3}$$

Since the parabola “smiles” and we are interested in the sections below the x-axis, we get

$$-\frac{1}{3}

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