## Thursday, October 15, 2009

### Period 4: Solving Inequalities Algebraically

Inequality deals with a range of values, rather than an exact number, so it is necessary to replace the equal sign in an equation with its supposed inequality (either >, <, >, or <).

For example, Solve for the following inequality:

(x + 10) (2x – 9) > 0

1. Break up both of the brackets into two separate “sections”, x +12>0 and 2x - 9 >0.
2. So far, we know that each individual section must exceed the value of 0, as this is the inequality. From this, we can isolate the x value to determine the roots.

x+ 12 > 0
x > -10

2x-9 >0
= 2x> 9
x > 9/2

Note that for section 2, the value of x can also be 0.

So now we know that the roots of the equation when (x + 10) (2x – 9) = 0
are x = -10, x = 9/2, and x = 0

1. We can then break up number line into three intervals in relation to the roots that we have found.

Because the lowest root is -10, we recognize in order for the graph to continue negatively, x must therefore be less than -10, so x <
-10.

Likewise, the highest root value is 9/2 so in order for the graph to continue positively, x must be bigger than 9/2, so 9/2 > x. We must also notice that the middle interval must (graphically) be larger than -10, yet smaller than 9/2, so, respectively, the inequality is represented by -10 <>

2. You can test out each interval by plugging it into the equation and determining which one is the solution.

For x<-10, test x = -11

(-11 + 10) (2 (-11) -9) = 31
because the value “31” is greater than “0”, x< -10 is a solution. For -10 <9/2,>x = 1
(1+10) (2(1)-9) = -77
because the value “-77” is less than “0”, -10 <>
solution. For x > 9/2, test x = 5
(5+10) (2(5)-9) = 135
because the value “135” is greater than “0”, x>9/2 is a solution.

3. Summarize the information found onto a table:
 Interval Factor x < -10 -10 x > 9/2 (x +10) ( - ) ( + ) ( + ) (2x – 9) ( - ) ( - ) ( + ) f(x) ( + ) ( - ) ( + )

4. The solution, shown on the number line, is x < -10 and x > 9/2

Note: Page 8 of our textbooks summarizes all possible intervals

Therefore, the solution is XE(-∞, -10] U[9/2, ∞)