Day: April 29, 2018

Week 11 – Solving Quadratic Inequalities

manroopt2015 on

This week in Pre-Calculus 11 we started the Solving Quadratic Inequalities unit. This week we learned about how to solve inequalities with one variable, graphing linear inequalities, and graphing quadratic inequalities.

How to Solve an Inequality with One Variable: To solve an inequality you have to isolate the variable in order to figure out the possible values of the variable, then you should test a number accordingly. When solving an inequality there are some rules that you need to remember. In order for the inequality to be true when you are diving by a negative number or multiplying by a negative number you MUST flip the inequality symbol.

NOT FLIPPING THE SIGN

Ex. \frac{1}{-2}x>2

x>(2)(-2)

x>-4 

TEST: x=0

\frac{1}{-2}\times0>2

0>2

                     

FLIPPING THE SIGN

Ex. \frac{1}{-2}x>2

x<2\times-2

x\prec-4

TEST: x=-5

\frac{1}{-2}\times\frac{-5}{1}>2

\frac{-5}{-2}>2

2.5>2

In the example above I showed how the inequality is not true unless you flip the sign when multiplying and dividing by a negative number. When I didn’t flip the sign, the inequality for x did not work which meant that it was false. The graph in the example represents the inequality. The shaded area represents all of the possible values for x in order to make the inequality true. The shaded area and the unshaded area is separated by a broken line this represents the < sign.

How to Solve a Quadratic Inequality: To solve a quadratic inequality you have to factor the inequality and find the values of x. After finding the possible values of x you can test points between the values of x using a number line.

Ex. x^2+12x+20>0

(x+10)(x+2)

x+10=0                  x+2=0

x=-10                                    x=-2

 

 

-10>x>-2

In the example above, I used a number line to solve the quadratic inequality. The first step I did was to find the zero’s of the quadratic function by factoring. Then I placed the zero’s on a number line and tested numbers between the zero’s. After testing the values using the original inequality I found out what numbers satisfied the inequality.