Lesson 4-6 Rational Equations and Partial Fractions

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Presentation transcript:

Lesson 4-6 Rational Equations and Partial Fractions Objective: To solve rational equations and inequalities To decompose a fraction into partial fractions

Rational Equations Rational Equation – has 1 or more rational expressions. Solve by multiplying each side by the LCD

Rational Equations To solve a rational equation: 1. Find the LCM of the denominators. 2. Clear denominators by multiplying both sides of the equation by the LCM. 3. Solve the resulting polynomial equation. 4. Check the solutions.

Find the LCM. Multiply by LCM = (x – 3). Solve for x. Check. Examples: 1. Solve: . LCM = x – 3. Find the LCM. 1 = x + 1 Multiply by LCM = (x – 3). x = 0 Solve for x. (0) Check. Substitute 0. Simplify. True. 2. Solve: . LCM = x(x – 1). Find the LCM. Multiply by LCM. x – 1 = 2x Simplify. x = –1 Solve. Examples: Solve

Check. x = 3 is not a solution since both sides would be undefined. Example: Solve: . x2 – 8x + 15 = (x – 3)(x – 5) Factor. The LCM is (x – 3)(x – 5). Original Equation. x(x – 5) = – 6 Polynomial Equation. Simplify. x2 – 5x + 6 = 0 Factor. (x – 2)(x – 3) = 0 Check. x = 2 is a solution. x = 2 or x = 3 Check. x = 3 is not a solution since both sides would be undefined. Example: Solve

Decomposing a fraction into Partial Fractions. Sometimes we need more tools to help with rational expressions… We will learn to perform a process known as partial fraction decomposition… To find partial fractions for an expression, we need to reverse the process of adding fractions.

To find the partial fractions, we start with The expressions are equal for all values of x so we have an identity. The identity will be important for finding the values of A and B.

To find the partial fractions, we start with Multiply by the LCD So, If we understand the cancelling, we can in future go straight to this line from the 1st line.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. Why should I choose x = 2 ? ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. What value would you substitute next ? ANS: x = - 1 so that the first term becomes 0.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. So,

Example 2 Express the following as 2 partial fractions. Solution: Let Multiply by :

Decomposing Fractions Decompose into partial fractions 2/(x+2) +4/(x-5)

Rational Inequalities To solve rational inequalities: Find the zeros and mark on a number line. Find any exclusions (restrictions) and mark on a number line. Test a value on each interval.

Rational Inequalities Solve Set to 0 LCD=15b Find the zeros

Rational Inequalities -1/15 Test b<-1/15 try b=-1 True Test -1/15<b<0 try b=-1/30 False Test b>0 try b=1 True

Rational Inequalities So

Rational Inequalities Solve -2<x<-1, -1<x<1,1<x<3, x>3

Sources www.collinsville.k12.ok.us/webpages/documents/HS- SparksR/PC4.6.ppt, Jan 11, 2014.