11-3: SOLVING RADICAL EQUATIONSEssential 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 +3 Add 3 to each side ( )2 = (7)2 Square both sides to

remove the radical x = 49

x

x

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

3x

3x

11-3: Solving Radical Equations YOUR TURN: Simplify each radical

25

81

38

7 12x

4 5a

2 6c

11-3: Solving Radical Equations 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?

8 2v h r

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

8 2v h r

22

30 8 2(18)

30 8 36

3.75 36

3.75 36

14.0625 36

50

8 8

36 36

.0625

h

h

h

h

h

h

11-3: Solving Radical Equations YOUR TURN

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.

67 ft

8 2v h r

11-3: Solving Radical Equations 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

2 2

3 2 6

3 2 6

3 2 6

2 2

2 8

4

n n

n n

n n

n n

n

n

11-3: Solving Radical Equations Your Turn

Solve Check your answer

x = 5

3 4 5 6x x

11-3: Solving Radical Equations 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 x2 = x + 6 Looks like a quadratic… x2 – x – 6 = 0 Subtract x & 6 from each

side (x – 3)(x + 2) = 0 Factor the quadratic x – 3 = 0 | x + 2 = 0 Set each side = 0 x = 3 | x = -2 Solve for x Check each solution back in the original

6x x 22

6x x

11-3: Solving Radical Equations Your Turn

Solve Find the real and extraneous solution

Real: y = 2 Extraneous: y = -1

Note: If both of the solutions are extraneous, we say there is no solution.

2y y

11-3: Solving Radical Equations Assignment

Worksheet #11-3 1 – 35, odds