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#### xibu

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« on: April 06, 2022, 06:26:01 pm »
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so I understand how to solve quadratic inequalities and quadratic inequalities with an unknown denominator of one variable.

But I can't seem to be able to do problems where an expression is in the denominator, I know that you multiply both sides by the square of the expression but get stuck at the point where you need to factor out the expression to get a quadratic.

for example:

sorry for the trivial question, any explanations will be appreciated.
« Last Edit: April 06, 2022, 06:34:06 pm by xibu »

#### 1729

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« Reply #1 on: April 06, 2022, 08:16:19 pm »
+1
so I understand how to solve quadratic inequalities and quadratic inequalities with an unknown denominator of one variable.

But I can't seem to be able to do problems where an expression is in the denominator, I know that you multiply both sides by the square of the expression but get stuck at the point where you need to factor out the expression to get a quadratic.
Generally speaking, when approaching inequalities involving familar graphs like the one in your example, or a parabola -- it might be easier to quickly sketch it and make sense out of what you are solving. Graphical Approach
For instance to solve $\frac{1}{x+3}<-5$. Sketch the actual hyperbola and the line $y=-5$ and label it's point of intersection. It should be easy to recognize that the solution to the inequality is $\left(-\frac{16}{5},\:-3\right)$ after finding that it the $\frac{1}{x+3}$ intersects $y=-5$ at $x=-\frac{16}{5}$.

Sketch of the graph
Below is a sketch of the graph where the green line is the asymptote  and blue line is $y=-5$ If you really wanted to turn the inequality into a quadratic inequality, you could do so as such.
1. Multiply the inequality $\frac{1}{x+3}<-5$ by $\left(x+3\right)^2$. This will yield to the following results:
$\Longrightarrow \frac{\left(x+3\right)^2}{x+3}<-5\left(x+3\right)^2$
$\Longrightarrow \left(x+3\right)<-5\left(x+3\right)^2$
$\Longrightarrow 0<-5\left(x+3\right)^2-\left(x+3\right)$
2. Try and factor the resulting expression by factoring out $\left(x+3\right)$ as such
$\Longrightarrow 0<-5\left(x+3\right)^2-\left(x+3\right)$
$\Longrightarrow 0<\left(x+3\right)\left[-5\left(x+3\right)-1\right]$
$\Longrightarrow 0<\left(x+3\right)\left(-5x-16\right)$
$\Longrightarrow 0<-\left(x+3\right)\left(5x+16\right)$

This is how we "factor out the expression to get the quadratic". From here you can solve as you would solve a quadratic inequality. HOWEVER, make sure you do not forget to consider that $x\ne -3$ because $\frac{1}{-3+3}$ is undefined. (ie. make sure you look back at your original function before considering whether it's inclusive or exclusive)

Hope this helps, and don't hesitate to ask any questions if you are confused! « Last Edit: April 06, 2022, 08:20:07 pm by 1729 »