11-3: SOLVING RADICAL EQUATIONS Essential Question: What is an extraneous solution, and how do you find one?

11-3: Solving Radical Equations  A radical equation is an equation that has a variable underneath a radical.  You solve radical equations by getting the radical alone on one side of the equation, then squaring both sides.  You should take your answers and substitute them back in the original problem to make sure your answers aren’t extraneous (we’ll get to that shortly)

11-3: Solving Radical Equations  Example 1a: Solving by Isolating the Radical  Solve each equation  - 3 = 4  +3 +3Add 3 to each side  ( ) 2 = (7) 2 Square both sides to remove the radical  x = 49

11-3: Solving Radical Equations  Example 1b: Solving by Isolating the Radical  Solve each equation  = 4  ( ) 2 = (4) 2 Square both sides to remove the radical  x – 3 = 16  +3 +3Add 3 to each side  x = 19

11-3: Solving Radical Equations YY OUR T URN : Simplify each radical

 Example 2: Real World Problem Solving  On a roller coaster, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation gives the velocity v in feet per second of a car at the top of the loop.  Suppose the loop has a radius of 18 feet. You want the car to have a velocity of 30 ft/s at the top of the loop. How high should the hill be?

11-3: Solving Radical Equations  Solve for h when v = 30 and r = 18  Substitute  Get the radical alone  Divide both sides by 8  Square both sides  Add 36 to both sides  Hill should be about 50 ft

11-3: Solving Radical Equations YY OUR T URN FFind the height of the hill when the velocity at the top of the loop is 35 ft/s and the radius of the loop is 24 ft. Round your answer to the nearest foot. 667 ft

 If you have a radical on each side of an equal sign, you can square both sides of the equation to solve.  Example 3: Solving With Radical Expressions on Both Sides  Solve Square each side Square and root cancel out Subtract n & +2 to each side Divide by 2

11-3: Solving Radical Equations YYour Turn SSolve CCheck your answer x = 5

 An extraneous solution is a solution that is found algebraically, but doesn’t satisfy the original equation. You can only find extraneous solutions by checking answers.  General Rule  Whenever there is an x inside a square root AND an x outside a square root, you’ll have to check for extraneous solutions.

11-3: Solving Radical Equations  Example 4: Identifying Extraneous Solutions   Square both sides  x 2 = x + 6Looks like a quadratic…  x 2 – x – 6 = 0Subtract x & 6 from each side  (x – 3)(x + 2) = 0Factor the quadratic  x – 3 = 0| x + 2 = 0Set each side = 0  x = 3| x = -2Solve for x  Check each solution back in the original

11-3: Solving Radical Equations YYour Turn SSolve FFind the real and extraneous solution Real: y = 2 Extraneous: y = -1 NNote: If both of the solutions are extraneous, we say there is no solution.

 Assignment  Worksheet #11-3  1 – 35, odds