Objective 1.4.1
Radical Equations

To solve an equation with a radical
2x −1 − 3 = 4 First isolate the radical
+3 +3 This means to get any terms not
under the square root on the other
2 2 side of the equal sign
Now square both sides
2x −1 = 7 You must square the whole side
NOT each term. A square
2 x − 1 = 49 Now solve for x
"undoes" or cancels a square root
+1 +1 You MUST check this answer
2 x = 50 2( 25) − 1 − 3 = 4
Since you squared both sides of the
equation, negatives disappear. It is
x = 25 4=4
possible to get anIt checks! doesn't work
answer that
when you plug it back in

Let's try another one:
1
( 2 x + 1) 3 +1 = 0
-1
First isolate the radical
Remember that the 1/3
-1 Now since it is same thing
power means the a 1/3
as a cube root.
power this means the
1 3 3
( 2 x + 1) 3= −1
same as a cube root so
cube both sides
2 x + 1 = −1 Now solve for x
-1 -1 Let's check this answer
2 x = −2 32( − 1) + 1 + 1 = 0
x = −1 0 = 0 It checks!

One more to see extraneous solution:
3 x + 1 =that − 3find algebraically but DOES isolated
a solution x you The radical is already NOT
make a true statement when you substitute it back
2
into the equation. 2 Square both sides
3x + 1 = x − 3 You must square the whole side
NOT each term.
This must be FOILed
3 x + 1 = x − 6 x + 9 You MUST check
2
Since you have a quadratic these answersan x
x −getxeverything on one side = 0 and see if
2
9 +8 = 0 equation (has 2
3 (8) + 1 = 1 − 3
term)
( x − 8)( x − 1) = 0
you can factor this 3 1 +1 = 8−3
x = 8, x = 1 It Doesn't work!
2=− checks!
5 = 5 2 Extraneous

Objective 1.4.2
Equations in
Quadratic Form
The "u" Substitution Method

x − 5x + 4 = 0
4 2
Before we solve the above equation, let's solve a
quadratic equation that we know how to solve.
u − 5u + 4 = 0
2
Factor
( u − 4)( u − 1) = 0 Set each factor = 0
and solve
u = 4, u = 1
Let's use this to solve the original equation by
letting u = x2.

x − 5x + 4 = 0
4 2
If u = x2 then square both sides and get u2 = x4.
Substitute u and u2 for x2 and x4.
u − 5u + 4 = 0
2 Factor
( u − 4)( u − 1) = 0 Set each factor = 0
and solve
u = 4, u = 1 x = 4, x = 1
2 2
Now that we've solved for u we have to re-substitute to
get x back. Remember u = x2 so let's substitute.
Solve for x by square-rooting both sides and don't forget
the ±
x = ±2, x = ±1

You can determine if an equation is of quadratic form
where you can use the "u" substitution method if you call
the middle variable and power u and then square it and
get the first term's variable and power. 1 1
(z ) = z
4
1 1 2 2
z 2
− 4z 4
+4=0
u − 4u + 4 = 0
2 So let u = z1/4 and get u2 = z1/2.
Substitute u and u2 for z1/4 and z1/2.
( u − 2)( u − 2) = 0 u=2
Factor & set each factor
= 0 and solve
1 Solve for z by raising both
u=z =2 4 1 sides to the 4th power
( z ) = ( 2)
4 4 4
z = 16

Let's try one more. Call the middle variable u and then
square it to see if you get the first term's variable.
(x ) = x
3 2 6
x − 7x − 8 = 0
6 3
So let u = x3 and get u2 = x6.
u − 7u − 8 = 0
2
Substitute u and u2 for x3 and x6.
( u − 8)( u + 1) = 0 Factor & set each factor = 0 and solve
u = 8, u = −1 x = 8, x = −1
3 3
Solve for x by taking the cube root of both sides
x = 2, x = −1