Solving Radical Equations and Inequalities 5-8

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Presentation transcript:

Solving Radical Equations and Inequalities 5-8 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

Objective Solve radical equations and inequalities.

Vocabulary radical equation radical inequality

A radical equation contains a variable within a radical A radical equation contains a variable within a radical. Recall that you can solve quadratic equations by taking the square root of both sides. Similarly, radical equations can be solved by raising both sides to a power.

Remember! For a square root, the index of the radical is 2.

Example 1A: Solving Equations Containing One Radical Solve each equation. Check Subtract 5. Simplify. Square both sides.  Simplify. Solve for x.

Example 1B: Solving Equations Containing One Radical Solve each equation. Check 3 7 7 5x - 7 84 = Divide by 7. 7 Simplify. Cube both sides.  Simplify. Solve for x.

Check It Out! Example 1a Solve the equation. Check Subtract 4. Simplify.  Square both sides. Simplify. Solve for x.

Check It Out! Example 1b Solve the equation. Check Cube both sides. Simplify. Solve for x. 

Check It Out! Example 1c Solve the equation. Check Divide by 6. Square both sides. Simplify.  Solve for x.

Example 2: Solving Equations Containing Two Radicals Solve Square both sides. 7x + 2 = 9(3x – 2) Simplify. 7x + 2 = 27x – 18 Distribute. 20 = 20x Solve for x. 1 = x

Example 2 Continued Check 3 3 

Check It Out! Example 2a Solve each equation. Square both sides. 8x + 6 = 9x Simplify. 6 = x Solve for x. Check 

Check It Out! Example 2b Solve each equation. Cube both sides. x + 6 = 8(x – 1) Simplify. x + 6 = 8x – 8 Distribute. 14 = 7x Solve for x. 2 = x Check 2 2 

Example 3 Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. –3x + 33 = 25 – 10x + x2 Simplify. 0 = x2 – 7x – 8 Write in standard form. 0 = (x – 8)(x + 1) Factor. x – 8 = 0 or x + 1 = 0 Solve for x. x = 8 or x = –1

Example 3 Continued Method 2 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. 3 –3 x 6 6  Because x = 8 is extraneous, the only solution is x = –1.

Check It Out! Example 3a Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. 2x + 14 = x2 + 6x + 9 0 = x2 + 4x – 5 Write in standard form. Factor. 0 = (x + 5)(x – 1) x + 5 = 0 or x – 1 = 0 Solve for x. x = –5 or x = 1

Check It Out! Example 3a Continued Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. 2 –2 x 4 4  Because x = –5 is extraneous, the only solution is x = 1.

Check It Out! Example 3b Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. –9x + 28 = x2 – 8x + 16 0 = x2 + x – 12 Write in standard form. Factor. 0 = (x + 4)(x – 3) x + 4 = 0 or x – 3 = 0 Solve for x. x = –4 or x = 3

Check It Out! Example 3b Continued Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions.  

To find a power, multiply the exponents. Remember!

Example 4A: Solving Equations with Rational Exponents Solve each equation. 1 3 (5x + 7) = 3 Write in radical form. Cube both sides. 5x + 7 = 27 Simplify. 5x = 20 Factor. x = 4 Solve for x.

Example 4B: Solving Equations with Rational Exponents 2x = (4x + 8) 1 2 Step 1 Solve for x. Raise both sides to the reciprocal power. (2x)2 = [(4x + 8) ]2 1 2 4x2 = 4x + 8 Simplify. 4x2 – 4x – 8 = 0 Write in standard form. 4(x2 – x – 2) = 0 Factor out the GCF, 4. 4(x – 2)(x + 1) = 0 Factor. 4 ≠ 0, x – 2 = 0 or x + 1 = 0 Solve for x. x = 2 or x = –1

Step 2 Use substitution to check for extraneous solutions. Example 4B Continued Step 2 Use substitution to check for extraneous solutions. 2x = (4x + 8) 1 2 2(2) (4(2) + 8) 4 16 4 4  2x = (4x + 8) 1 2 2(–1) (4(–1) + 8) –2 4 –2 2 x The only solution is x = 2.

Check It Out! Example 4a Solve each equation. (x + 5) = 3 1 3 (x + 5) = 3 Write in radical form. Cube both sides. x + 5 = 27 Simplify. x = 22 Solve for x.

Raise both sides to the reciprocal power. Check It Out! Example 4b 1 2 (2x + 15) = x Step 1 Solve for x. [(2x + 15) ]2 = (x)2 1 2 Raise both sides to the reciprocal power. 2x + 15 = x2 Simplify. x2 – 2x – 15 = 0 Write in standard form. (x – 5)(x + 3) = 0 Factor. x – 5 = 0 or x + 3 = 0 Solve for x. x = 5 or x = –3

Check It Out! Example 4b Continued Step 2 Use substitution to check for extraneous solutions. (2(5) + 15) 5 1 2 (2x + 15) = x (10 + 15) 5  5 5 (2(–3) + 15) –3 1 2 (–6 + 15) –3 x (2x + 15) = x 3 –3 The only solution is x = 5.