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# 11.5

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### 11.5

1. 1. Vocabulary  Rational Equation - an equation that contains one or more rational expressions.  When both sides of the equation are single rational expressions, you can solve the equation by using the cross products property.  Extraneous Solutions – A solution of a transformed equation that is not a solution of the original equation.
2. 2. Example 1 Use the cross products property 6 x Solve = . Check your solution. 2 x +4 6 x = 2 x +4 Write original equation. 12 = x2 + 4x Cross products property 0 = x2 + 4x – 12 Subtract 12 from each side. 0 = (x + 6) (x – 2) Factor polynomial. x + 6 = 0 or x – 2 = 0 x = – 6 or x = 2 Zero-product property Solve for x.
3. 3. Example 1 ANSWER CHECK Use the cross products property The solutions are – 6 and 2. If x = – 6: If x = 2: 6 ? –6 = –6 + 4 2 6 ? 2 = 2 +4 2 –3 = –3 1= 1
4. 4. Example 2 Solve Multiply by the LCD x x – 2 2 1 + = 5 x – 2 . Check your solution. Multiply each side of the equation by the LCD to transform the rational equation into a polynomial equation. x x – 2 x x – 2 • 5 (x – 2 ) + 1 5 2 1 + = 5 • 5 (x – 2 ) = Write original equation. x – 2 2 x – 2 • 5 (x – 2 ) Multiply by LCD, 5 ( x – 2).
5. 5. Example 2 Multiply by the LCD x • 5 (x – 2 ) x – 2 5 (x – 2 ) 2 • 5 (x – 2 ) = + x – 2 5 Multiply, then divide out common factors. 5x + x – 2 = 10 Simplify. 6x – 2 = 10 Combine like terms. 2 Solve for x. x = The solution appears to be 2, but the expressions and 2 x – 2 are undefined when x = 2. So, 2 is an extraneous solution. ANSWER The rational equation has no solution. x x – 2
6. 6. Example 3 Solve 3 x – 7 Factor to find the LCD + 1 = 8 x2 – 9x + 14 . Check your solution. Write each denominator in factored form. The LCD is ( x – 2 ) ( x – 7 ). 3 8 x – 7 3 x – 7 + 1 = (x – 2 ) (x – 7 ) • (x – 2 ) (x – 7 ) + 1 • (x – 2 ) (x – 7 ) 8 = (x – 2 ) (x – 7 ) • (x – 2 ) (x – 7 )