Objective Solve radical equations A radical equation contains

Objective Solve radical equations.

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

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

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

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

Check It Out! Example 1 c 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. 7 x + 2 = 9(3 x – 2) Simplify. 7 x + 2 = 27 x – 18 Distribute. 20 = 20 x 1=x Solve for x.

Example 2 Continued Check 3 3

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

Raising each side of an equation to an even power may introduce extraneous solutions. Helpful Hint You can use the intersect feature on a graphing calculator to find the point where the two curves intersect.

Example 3: Solving Equations with Extraneous Solutions Solve . Method 1 Use a graphing calculator. Let Y 1 = and Y 2 = 5 – x. The graphs intersect in only one point, so there is exactly one solution. The solution is x = – 1

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

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.