Rational Function, Extraneous solution

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Presentation transcript:

Rational Function, Extraneous solution Objectives: Be able to determine if an equation is a rational equation. Be able to solve various rational equations and exclude any extraneous solutions. Critical Vocabulary: Rational Function, Extraneous solution

Rational equation An equation that contains one or more rational expressions Solve a rational equation by multiplying both sides of the equal sign by the LCD If both sides of the equal sign are rational, you can solve be cross multiplying because it will be a proportion Least common multiple: smallest positive whole number exactly divisible by the given numbers

Formal Definition: A rational Function is a ratio of two I. Rational Functions A rational equation is an equation that contains rational expressions (x in the denominator) Formal Definition: A rational Function is a ratio of two polynomials written in the form No, that would be a linear equation. Looks like you got it. I think I get it. I bet this is a rational equation. What is a Rational Equation? In Simple terms, it’s a fraction. So, would this equation be rational? So, what is a rational then?

This is not an extraneous solution either. II. Solving Rational Equations b. Solving a Rational: By Finding LCM (Denominator) 1. What can x not be? Multiply by LCD Distribute 5x –6 =4 5x = 10 This is not an extraneous solution either. Solution: x = 2 No….really?

II. Solving Rational Equations a. Solving a Rational: Cross Multiplication (Proportion) 5. First determine what “x” can’t be (6)(x + 5) = (x +5)(x + 3) Cross Multiply 6x+30 = x2 + 8x + 15 Make it equal 0 0 = x2 +2x-15 What are talking about? What is an extraneous solution? 0= (x + 5)(x – 3) x = -5 x = 3 Solution: x = 3 That’s where your solution is one of the values that “x” can’t be. -5 is an extraneous solution.

Both of these solutions look good. II. Solving Rational Equations b. Solving a Rational: By Finding LCM (Denominator) 9. What can x not be? Multiply by LCD Distribute 5 =6x -x2 x2 –6x+5 =0 Make it equal 0 (x - 1)(x - 5)=0 Both of these solutions look good. x=1 x = 5 right Solution: x = 1,5