Unit 4 Radicals
SOLVING RADICALS
Steps to solve radical equations
 A radical equation is a equation with a variable
under the radical sign!
STEPS:
1) Isolate radical
2) get rid of radical
3) solve
4) check
“Square a square root”
“Cube a cube root”
ETC…
Solve each radical equation
Find all solutions.
3  2x  3  8
Answer:
X = 14
5 3 x  1  35
Answer:
X = 342
SOLVE
 Remember rational exponents represents radicals
 To remove rational exponents raise each side the
reciprocal power
3
5
3( x  1)  1  25
Exponent like
this is the same
as a radical.
POWER
INDEX
3 ( x  1)  1  25
5
Answer: X=31
3
Solve
 Solve
x 7 5  x
When X’s on both
sides of equation
solve by graphing!
To solve by graphing:
1)Graph
2) Find x-intercepts
Answer:
X = -3
Word problem
 You can model time, t in seconds, an object takes to
reach the ground falling from height H, in meters by
the equation:
2H
t(H ) 
9.81
 If an object takes 7 seconds to fall to the ground,
what was the initial height?
 If an object takes 10 seconds?
 How long will it take an object to hit the ground from
a initial height of 100ft?
Objects falling hammer vs. feather
practice
 Pg 394 #9-21 odd 26, 29, 32
Solving radicals
x  2x  6
VIDEO SOLUTION:
The solution/root/zero to a function is where the graph crosses the x-axis
Graph and find the intersections = solutions
Word problem

V
d  23
For a meteor crater in Arizona, the formula relates
0.3
the diameter d of the rim(in meters) to the volume V (in
cubic meters). What is the volume of the crater if the
diameter is 1200 meters
V
1200  2
0.3
3

V
600 
0.3
Solve by
By hand will
graphing or
be easier
by hand?
3
6003 
V
0.3
V  0.3  6003  64,800,000
Cube both
sides
Applying radicals to problems
 Radius of a sphere can be measured using the
formula: r  3 3V
4
 Find the radius if the volume is 250 ft3
 Find the volume if the radius is 6 (K.A. video)
 Pg. 365 #35, 36
Word problem
 You can model the population of a Texas town
between the years 1970 & 2005 by the function
below. x is the year. In what year did the town have a
population of 250,000?
p ( x)  75,0003 x  1950
250,000  75,0003 x  1950
10 3
 x  1950
3
37.037  x 1950
x  1987.037
Practice
 Pg 395 #24,25,45*,60
 Pg 418 #24, 54