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
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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