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Bell Quiz 6-6 Evaluate the expression. 64 3 pts 1. 3 pts 2. (x – 5)2 Solve the equation. 4 pts 3. x2 + 6x + 9 = x + 45 10 pts possible Section 6-6 1 6-6 Chapter 6: Rational Exponents and Radical Functions Section 6-6 2 6-6 Chapter 6: Rational Exponents and Radical Functions Section 6-6 3 Definition Radical Equations Equations with radicals that have variables in their radicands are called radical equations Example: Section 6-6 4 Key Concept Solving Radical Equations To solve a radical equation, follow these steps: Step 1: Isolate the radical on one side of the equation (if necessary). Step 2: Raise each side of the equation to the same power to eliminate the radical. Step 3: Solve the equation. Section 6-6 5 EXAMPLE 1 Solve a radical equation 3 Solve √ 2x+7 = 3. Section 6-6 6 EXAMPLE 1 Solve a radical equation 3 Solve √ 2x+7 = 3. 3 √ 2x+7 = 3 ( √3 2x+7 )3 = 33 2x+7 = 27 2x = 20 x = 10 Write original equation. Cube each side to eliminate the radical. Simplify. Subtract 7 from each side. Divide each side by 2. Section 6-6 7 EXAMPLE 1 Solve a radical equation CHECK Check x = 10 in the original equation. 3 √ 2(10)+7 =? 3 3 √ 27 =? 3 3 = 3 Substitute 10 for x. Simplify. Solution checks. Section 6-6 8 GUIDED PRACTICE for Example 1 Solve equation. Check your solution. 1. 3√ x – 9 = –1 x = 512 2. (√ x+25 ) = 4 x = –9 Section 6-6 3. (23√ x – 3 ) = 4 x = 11 9 EXAMPLE 2 Solve a radical equation given a function Section 6-6 10 Rational Exponents When an equation contains a power with a rational exponent a.k.a fractional exponent Step 1: Isolate the power. Step 2: Raise each side of the equation to the reciprocal of the exponent Step 3: Solve the equation. Section 6-6 11 EXAMPLE 3 Standardized Test Practice Section 6-6 12 EXAMPLE 3 Standardized Test Practice SOLUTION 4x2/3 = 36 x2/3 = 9 (x2/3)3/2 = 93/2 x = 27 Write original equation. Divide each side by 4. 3 Raise each side to the power 2 . Simplify. ANSWER The correct answer is D. Section 6-6 13 EXAMPLE 4 Solve an equation with a rational exponent Solve (x + 2)3/4 – 1 = 7. Section 6-6 14 EXAMPLE 4 Solve an equation with a rational exponent Solve (x + 2)3/4 – 1 = 7. (x + 2)3/4 – 1 = 7 (x + 2)3/4 = 8 (x + 2)3/4 4/3 = 8 4/3 Write original equation. Add 1 to each side. Raise each side to the power 4 . 3 x + 2 = (8 1/3)4 Apply properties of exponents. x + 2 = 24 Simplify. x + 2 = 16 Simplify. x = 14 Subtract 2 from each side. Section 6-6 15 for Examples 3 and 4 GUIDED PRACTICE Solve the equation. 4. 3x3/2 = 375 5. 6. –2x3/4 = –16 x = 25 x = 16 – 2 x1/5 = –2 x = 243 3 Section 6-6 16 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. 7. (x + 3)5/2 = 32 8. (x – 5)5/3 = 243 9. (x + 2)2/3 +3 = 7 x=1 x = 32 x = –10 or 6 Section 6-6 17 Extraneous Solutions Raising each side of an equation to the same power may introduce extraneous solutions. a.k.a It will create extra answers that are not answers to the original problem. Always check solutions in original equation. Section 6-6 18 EXAMPLE 5 Solve an equation with an extraneous solution Solve x + 1 = √7x + 15 . Check for extraneous solutions. Section 6-6 19 EXAMPLE 5 Solve an equation with an extraneous solution Solve x + 1 = √7x + 15 x + 1 = √ 7x + 15 Write original equation. (x + 1)2 = (√7x + 15)2 Square each side. Expand left side and simplify right side. x2 + 2x + 1 = 7x + 15 x2 – 5x – 14 = 0 Write in standard form. Factor. (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2 Zero-product property Solve for x. Section 6-6 20 EXAMPLE 5 Solve an equation with an extraneous solution CHECK Check x = 7 in the original equation. Check x = –2 in the original equation. x + 1 = √7x + 15 x + 1 = √7x + 15 ? ? –2 + 1 = √ 7(–2) + 15 7 + 1 = √ 7(7) + 15 8 = √ 64 ? –1 = √ 1 8=8 –1 =/ 1 ANSWER ? The only solution is 7. (The apparent solution –2 is extraneous.) Section 6-6 21 GUIDED PRACTICE for Examples 5 Solve the equation. Check for extraneous solutions 10. − = 11. x = 1, ¼ 10x + 9 = x + 3 x = 0,4 Section 6-6 22 HOMEWORK Sec 6-6 (pg 456) 3-21 every 3rd, 23-25, 34-38 Section 6-6 23