Transcript Using the Squaring Property with a Radical on Each Side

Entry Task
• You are a passenger in a car. You are using a cell phone that connects
with the tower shown. The tower has an effective range of 6 miles. If
you look out the window you see the tower and estimate that it is 3
miles from you. How many miles do you have to finish your call?
• 5.19 miles
6 miles
3 miles
6.5 Solving Square Root and
Other Radical Equations
x 3  4
3 x  1  4 x  1  28
Learning Target:
I can solve square root and other
radical equations
A radical equation
is an equation
that contains a radical.
x 3  4
The goal in solving
radical equations
is the same as the goal
in solving most equations.
We need to isolate
the variable.
x 3  4
Solving a Radical Equation.
• Use the following steps when solving an equation with
radicals.
• Step 1
Isolate a radical. Arrange the terms so that
a radical is isolated on one side of the
equation.
Step 2
Square both sides.
Step 3
Combine like terms.
Step 4
Repeat Steps 1-3 if there is still a term with a
radical.
Step 5
Solve the equation. Find all proposed solutions.
Step 6
Check all proposed solutions in the original
equation.
We need to square the
radical expression.
x 3  4
What we do to one side,
we have to do to the other

x  3  4
2
2
Now we need to simplify:


x  3
2
2
x  3  4
x  3  16
x  19
2
Remember,
 n
2
n
no matter what
n is.
(Even if n is an expression)
Solve for x:
3 x  3  4 x  3  28
Step 1.
Step 2.
Step 3.
Step 4.
Simplify the expression:
Isolate the radical.
Square both sides.
Solve the equation.
3 x  3  4 x  3  28

7 x  3  28
x3 4
2
2
x  3  4
x  3  16
x  13
Solve for x:
2 5 x  5  5 5 x  5  15
 3 5 x  5  15
5x  5  5
5 x  5  25
5 x  20
x4
Try this one:
Check for extraneous solutions
EXAMPLE 1
Using the Squaring
Property of Equality
• Solve.
Solution:

9 x  4
9 x

2
4
2
9  x  16
9  x  9  16  9
x  7
x  7
7
EXAMPLE 2
•Solve.
Solution:
Using the Squaring Property
with a Radical on Each Side
3x  9  2 x

3x  9
  2 x 
2
2
3x  9  4x
3x  9  3x  4x  3x
x 9
9
EXAMPLE 3
Using the Squaring Property
when One Side Has Two Terms
• Solve 2 x  1  10 x  9.
 2 x  1   10 x  9 
4 x2  4 x  1  10 x  9  10 x  9  10 x  9
2
Solution:
2
4 x 2  14 x  8  0
 2x  1 2x  8  0
2x 1  0
or
2x  8  0
x4
1
x
2
After we check our work the solution set is {4}.
EXAMPLE
x  15  3  x
2
16  15  3  16
2
x  15 (3  x )
x  15  (3  x )(3  x)
1 3 4
1  1
x  15  9  6 x  x
NO SOLUTION
Since 16 doesn’t plug in
as a solution.
 15  9  6 x
 24  6 x
4 x
16  x
Let’s Double
Check that this
Note: You will
get Extraneous
Solutions from
time to time –
Solve
Solve
Can graphing calculators help?
SURE!
x  x2
1.
2.
3.
4.
Input x for Y1
Input x-2 for Y2
Graph
Find the points of
intersection
One Solution at (4,
2)
To see if this is extraneous or not, plug
the x value back into the equation. Does it
Assignment pg 395
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