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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.

Final Answer


-\frac{1}{2}<m<\frac{3}{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}<m<\frac{3}{2}

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