Inequalities – Quadratic equation with a parameter – Exercise 5725

Exercise 

Given the parabolic equation:

$y=(m+6)x^2-(3m+2)x+2$

For which values of parameter m, the parabola is on the x-axis or above it.

Final Answer

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Solution

Given the parabolic equation:

$y=(m+6)x^2-(3m+2)x+2$

We want to find out when:

$y\geq 0$

Or

$(m+6)x^2-(3m+2)x+2\geq 0$

Two conditions are required:

1. The parabola is “smiling” – the coefficient of the squared expression is greater than zero:

$m+6>0$

$m>-6$

2. The quadratic equation do not have a solution, meaning:

$\Delta<0$

Because in such a situation the whole parabola is always above the x-axis.

The coefficients of the quadratic equation are

$a=m+6, b=3m+2, c=2$

Putting the coefficients in the Delta formula (in the quadratic formula) gives us

$\Delta=b^2-4ac=$

$={(3m+2)}^2-4\cdot (m+6)\cdot 2=$

$=9m^2+12m+4-8m-48=$

$=9m^2+4m-44$

And we want it to take place:

$9m^2+4m-44<0$

It is a square inequality. Its coefficients are:

$a=9, b=4, c=-44$

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

$m_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 9\cdot (-44)}}{2\cdot 9}=$

$=\frac{-4\pm \sqrt{1600}}{18}=$

$=\frac{-4\pm 40}{18}$

Hence, we get the solutions:

$x_1=\frac{-4+ 40}{18}=2$

$x_2=\frac{-4- 40}{18}=\frac{-44}{18}=-2\frac{4}{9}$

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

$-2\frac{4}{9}\leq m\leq 2$

We combine (“and”) both results (of both conditions) and we still get that the final answer is

$-2\frac{4}{9}\leq m\leq 2$

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