Download Solving Radical Equations

Aim: How do we solve radical equations?
Do Now: Describe the steps for solving:
x2 – 80 = 0
add 80 to both sides
x2 = 80
take square root of both sides
x 2   80
x   80
simplify
Describe the reverse process
x   80
2


2
x   80 square both sides
x2 = 80
x2 – 80 = 0 subtract 80 from both sides
How do we solve?
solve by first squaring
x  x 4
both sides.
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Perfect Squares
12
144
121
100
81
11
10
9
8
64
7
49
6
36
5
25
4
16
3
9
2
1
4
1
1
2
3
4
5
Aim: Radical Equations
6
7
8
9 10 11 12
Course: Alg. 2 & Trigonometry
Simplifying Radicals
KEY: Find 2 factors for the radicand - one
of which is the largest perfect square
possible
50  25  2  25  2
 5 2  5 2
Multiplying Radicals
x a  y b  xy ab
n
n
n
ex . 3 6  5 2  3  5 6  2
 15 12  15 4  3  30 3
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Dividing Radicals
n
n
ex .
a n a

b
b
72
72

 93
8
8
If quotient is not a perfect square
you must simplify the radicand.
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Adding/Subtracting Radicals
•Must have same radicand and index
•Add or subtract coefficients and combine
result with the common radical
Coefficient
3 3 5 3 8 3
Common
Radical
Combined
Result
Unlike radicals must first be simplified
to obtain like radicals
(same radicand-same index), if possible.
ex. 2 90  3 40
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Solving Radical Equations
Solve and check:
x  x 4
Isolate the radical:
x  x 4
(already done)
Square each2 side:
2
2
x

x


x

4


x
 8x  16
 
Solve the derived
equation:
x2 – 9x + 16 = 0
b  b2  4ac
use quadratic
x
formula:
2a
9  17
x
2
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Solving Radical Equations
Solve and check:
x 2  5
Isolate the radical:
x 2  5
(already done)
2
2
Square each side:
( x  2)  5
Solve the derived
equation:
x – 2 = 25
x = 27
Check:
27  2  5
alternate:
(x – 2)1/2 = 5
25  5
5=5
[(x – 2)1/2]2 = 52
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Extraneous Roots
Solve and check:
2y  1  7  4
Isolate the radical:
2y  1  3
Square each side:
Solve the derived
equation:
Check:
2
( 2y  1)  (3)
2y – 1 = 9
2y = 10
y=5
2(5)  1  7  4
974
y = 5 is an extraneous root; 3 + 7 = 4
there is no solution!
Aim: Radical Equations
2
?
Course: Alg. 2 & Trigonometry
Solving Radical Equations
Solve and check:
x 1 x 5
Isolate the radical:
x1 x 5
2
Square each side: (x  1)  ( x  5)
x2 – 2x + 1 = x + 5
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x=4
x = -1
Solve the derived
equation:
Check
each
root:
4  1 4  5
1  1  1  5
4  1 9
1  1  4
4=4
Aim: Radical Equations
2
?
x = -1 is an extraneous root
Course: Alg. 2 & Trigonometry
Solving Radical Equations
Solve and check: 3 x  2  2 x  8  0
?
3 x2  2 x8

  2
Square each side: 3 x  2
Solve the derived
equation:
2
x8

2
32(x – 2) = 22(x + 8)
9(x – 2) = 4(x + 8)
9x – 18 = 4x + 32
x = 10
x = 10 checks out as the solution
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Model Problem
The radical function h(x)  0.4 3 x is an
approximation of the height in meters of
a female giraffe using her weight x in
kilograms. Find the heights of female giraffes
with weights of 500 kg. and 545 kg.
Evaluate for 500: h(500)  0.4 3 500
 3.17 m.
Evaluate for 545: h(545)  0.4 3 545
 3.27 m.
Aim: Radical Equations
Course: Alg. 2 & Trigonometry
Model Problem
2 2r gives the time T in
F a body with mass 0.5 kg
The equation T 
seconds it takes
to complete one orbit of radius r meters. The force F
in newtons pulls the body toward the center of the
orbit.
a. It takes 2 s for an object to make one revolution
with a force of 10 N (newtons). Find the radius of the
orbit.
b. Find the radius of the orbit if the force is 160 N
and T = 2.
a. T 
2 2r
F
2
20   2r
Aim: Radical Equations
2 2r
2 2 r
2
4
40  2 r
10
10
20
2  r  2.03

Course: Alg. 2 & Trigonometry