aim: how do we solve radical equations?
DESCRIPTION
x 2 = 80. x 2 = 80. x 2 – 80 = 0. Aim: How do we solve radical equations?. x 2 – 80 = 0. add 80 to both sides. Do Now: Describe the steps for solving:. take square root of both sides. simplify. Describe the reverse process. square both sides. subtract 80 from both sides. - PowerPoint PPT PresentationTRANSCRIPT
Aim: Radical Equations Course: Alg. 2 & Trigonometry
x2 80 2
x x 4
Aim: How do we solve radical equations?
Do Now: Describe the steps for solving: x2 – 80 = 0
x2 = 80
x2 80
x 80
Describe the reverse process
add 80 to both sides
take square root of both sides
simplify
x2 – 80 = 0
x 80
subtract 80 from both sides
square both sides
x2 = 80
solve by first squaringboth sides.
How do we solve?
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Perfect Squares
12
12 144
11
11 121
100
10
10
9
9 81
8
8 64
7
7 49
6
6 36
5
5 25
4
4 16
3
3 9
42
21
1 1
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Simplifying Radicals
KEY: Find 2 factors for the radicand - one of which is the largest perfect square
possible
50 225 225 25 25
n n nx a y b xy ab
26532563. ex
1215 3303415
Multiplying Radicals
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Dividing Radicals
an
bn
a
bn
398
72
8
72. ex
If quotient is not a perfect squareyou 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
Combined Result
Coefficient
3533 Common Radical
38
Unlike radicals must first be simplified to obtain like radicals
(same radicand-same index), if possible.
403902 ex.
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Solve and check:
Isolate the radical:(already done)
Solving Radical Equations
Square each side:
Solve the derived equation:
x x 4
x x 4
x 2
x x 4 2 x2 8x 16
x2 – 9x + 16 = 0
x b b2 4ac
2a
x 9 17
2
use quadratic formula:
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Solve and check:
Isolate the radical:(already done)
Solving Radical Equations
x 2 5
x 2 5
Square each side:
( x 2)2 52
Solve the derived equation:
x – 2 = 25
x = 27
Check:
27 2 5
25 5
5 = 5(x – 2)1/2 = 5
[(x – 2)1/2]2 = 52
alternate:
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Solve and check:
Extraneous Roots
2y 1 7 4
Isolate the radical:
Square each side:
( 2y 1)2 ( 3)2
Solve the derived equation:
2y – 1 = 9
y = 5
Check:
2y 1 3
2y = 10
2(5) 1 7 4
9 7 4
3 + 7 = 4 ?y = 5 is an extraneous root;there is no solution!
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Solve and check:
Solving Radical Equations
x 1 x 5
Isolate the radical:
Square each side:
Solve the derived equation:
Check each root:
x = -1 is an extraneous root
x 1 x 5
(x 1)2 ( x 5)2
x2 – 2x + 1 = x + 5x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0 x = 4 x = -1
4 1 4 5 1 1 1 5
4 = 4
4 1 9 1 1 4 ?
Aim: Radical Equations Course: Alg. 2 & Trigonometry
?
Solving Radical Equations
3 x 2 2 x 8 0Solve and check:
Square each side:
Solve the derived equation:
3 x 2 2 x 8
3 x 2 2
2 x 8 2
32(x – 2) = 22(x + 8)9(x – 2) = 4(x + 8)
9x – 18 = 4x + 32x = 10
x = 10 checks out as the solution
Aim: Radical Equations Course: Alg. 2 & Trigonometry
Evaluate for 500:
Evaluate for 545:
Model Problem
The radical function is an approximation of the height in meters ofa female giraffe using her weight x in kilograms. Find the heights of female giraffeswith weights of 500 kg. and 545 kg.
h(x) 0.4 x3
h(500) 0.4 5003 3.17 m.
h(545) 0.4 5453 3.27m.
Aim: Radical Equations Course: Alg. 2 & Trigonometry
The equation gives the time T in seconds it takes a body with mass 0.5 kg 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.
T 2 2r
F
Model Problem
T 2 2r
Fa.
2 2 2r
10
4 2 2r
10
40 2 2r
20 2r
20
2 r 2.03