1. Add: 5 x2 – 1 + 2x x2 + 5x – 6 ANSWERS 2x2 +7x + 30

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1. Add: 5 x2 – 1 + 2x x2 + 5x – 6 ANSWERS 2x2 +7x + 30 (x +1)(x – 1)(x + 6)

( ) EXAMPLE 1 Solve a rational equation with two solutions 1 – 8 x – 5 = 3 x Solve: 1 – 8 x – 5 = 3 x Write original equation. x(x – 5) 1– 8 x – 5 ( ) = 3 x Multiply each side by the LCD, x(x–5). x(x –5) – 8x = 3(x – 5) Simplify. x2 – 5x – 8x = 3x – 15 Simplify. x2 – 16x +15 = 0 Write in standard form. (x – 1)(x – 15) = 0 Factor. x = 1 or x = 15 Zero product property

EXAMPLE 1 Solve a rational equation with two solutions ANSWER The solutions are 1 and 15. Check these in the original equation.

GUIDED PRACTICE Solve the equation by using the LCD. Check for extraneous solutions. 7 2 + 3 x = 3 1. x = – 6 ANSWER 2 x + 4 3 = 2 2. x = 3 ANSWER 3 7 + 8 x = 1 3. x = 14 ANSWER

EXAMPLE 2 Check for extraneous solutions 6 x – 3 = 8x2 x2 – 9 – 4x x + 3 Solve: SOLUTION Write each denominator in factored form. The LCD is (x + 3)(x – 3). = 8x2 (x + 3)(x – 3) – 4x x + 3 6 x –3 x + 3 (x + 3)(x – 3) 6 x –3 = 8x2 –(x + 3)(x – 3) 4x 6(x + 3) = 8x2 – 4x(x – 3) 6x + 18 = 8x2 – 4x2 + 12x

EXAMPLE 2 Check for extraneous solutions 0 = 4x2 + 6x –18 0 = 2x2 + 3x – 9 0 = (2x – 3)(x + 3) 2x – 3 = 0 or x + 3 = 0 x = 3 2 or x = –3 You can use algebra or a graph to check whether either of the two solutions is extraneous.

EXAMPLE 2 Check for extraneous solutions Algebra 3 2 The solution, checks but the apparent solution –3 is extraneous, because substituting it in the equation results in division by zero, which is undefined. 6 –3 –3 = 4(–3) –3 +3 8(– 3)2 (–3)2 – 9 – Division by zero is undefined.

GUIDED PRACTICE Solve the equation by using the LCD. Check for extraneous solutions. 4. 3 2 + 4 x –1 x +1 = x = – 3 ANSWER 3x x +1 – 5 2x = 3 5. ANSWER x = 2 or x = 2 3 – 5x x –2 = 7 + 10 6. no solution ANSWER