Transcript Simplifying Radicals

Lesson 10-4 Warm-Up
ALGEBRA 1
“Solving Radical
Equations” (10-4)
What is a “radical Radical Equation: an equation that has a variable in the radicand
equation”?
Example:
How do you
solve a radical
equation?
To solve a radical equation:
1. Isolate the radical containing the variable (get it by itself), and
2. Square both sides of the equation to get rid of the radical [To get rid of a
radical, or square root, square it:
].
Example: Solve
.
+3 +3
Add 3 to both sides to isolate the radical
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
Solve each equation. Check your answers.
a.
x–5 =4
x–5 =4
x=9
(
x)2 = 92
Isolate the radical on the left side of the equation.
Square each side.
x = 81
Check:
x–5 = 4
81 – 5
4
9–5
Substitute 81 for x.
4
4=4
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
(continued)
x–5 =4
b.
(
x – 5)2 = 42
x – 5 = 16
Square each side.
Solve for x.
x = 21
Check:
x–5 = 4
21– 5 = 4
16 = 4
Substitute 21 for x.
4=4
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
On a roller coaster ride, 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 v = 8 h – 2r gives the velocity v in
feet per second of a car at the top of the loop.
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
(continued)
The loop on a roller coaster ride has a radius of 18 ft.
Your car has a velocity of 120 ft/s at the top of the loop.
How high is the hill of the loop you have just come
down before going into the loop?
Solve v = 8
120 = 8
120
= 8
8
15 =
h – 2r for h when v = 120 and r = 18.
h – 2(18)
Substitute 120 for v and 18 for r.
h – 2(18)
8
h – 36
(15)2 = ( h – 36)2
225 = h – 36
261 = h
The hill is 261 ft high.
Divide each side by 8 to isolate the radical.
Simplify.
Square both sides.
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
Solve
(
3x – 4)2 = (
3x – 4 =
2x + 3)2
3x – 4 = 2x + 3
3x = 2x + 7
x=7
Check:
3x – 4 =
3(7) – 4
17 =
2x + 3.
Square both sides.
Simplify.
Add 4 to each side.
Subtract 2x from each side.
2x + 3
2(7) + 3
Substitute 7 for x.
17
The solution is 7.
ALGEBRA 1
“Solving Radical
Equations” (10-4)
What is an
“extraneous
solution”?
Extraneous Solution: a solution that does not solve or satisfy the original equation
(in other words, it isn’t a solution, because it doesn’t make the equation a true
statement) – Extraneous solutions can sometimes occur when you square both sides
of an equation in order to get rid of a radical.
Example:
-2 is an extraneous solution, because it doesn’t work in the original equation.
Note: Always be sure to check all solutions by substituting them back into the original
equation to make sure there they aren’t extraneous.
Example:
ALGEBRA 1
“Solving Radical
Equations” (10-4)
Is it possible for
an equation to
have no
solutions?
It is possible for all of the solutions of an equation to be extraneous. This would
mean that the equation has no solutions.
Example:
x = 2 does not solve the original equation. Therefore,
solutions
has no
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
Solve x =
(x)2 = (
x + 12.
x + 12)2
Square both sides.
x2 = x + 12
x2 – x – 12 = 0
Simplify.
(x – 4)(x + 3) = 0
(x – 4) = 0 or (x + 3) = 0
x = 4 or x = –3
Check:
x=
4
4 = 4
Solve the quadratic equation by factoring.
Use the Zero–Product Property.
Solve for x.
x + 12
4 + 12
–3
–3 + 12
–3 ≠ 3
The solution to the original equation is 4.
The value –3 is an extraneous solution.
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Additional Examples
Solve
3x + 8 = 2.
3x = –6
(
3x)2 = (–6)2
Subtract 8 from each side.
Square both sides.
3x = 36
x = 12
Check:
3x + 8 = 2
3(12) + 8
2
36 + 8
2
6 +8≠2
Substitute 12 for x.
x = 12 does not solve the original equation.
3x + 8 = 2 has no solution.
ALGEBRA 1
Solving Radical Equations
LESSON 10-4
Lesson Quiz
Solve each radical equation.
25
1.
7x – 3 = 4
3.
2x + 7 =
5.
3x + 4 + 5 = 3
2.
7
5x – 8 5
3x – 2 =
4. x =
2x + 8
x+2
2
4
no solution
ALGEBRA 1