Post by: xibu on April 06, 2022, 06:26:01 pm
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.
Post by: 1729 on April 06, 2022, 08:16:19 pm
so I understand how to solve quadratic inequalities and quadratic inequalities with an unknown denominator of one variable.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. :)
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.
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 \(y=-5\)
(https://i.imgur.com/IcKiihR.png)
(https://i.imgur.com/IcKiihR.png)
Quadratic Approach
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! :)