Download Quad u Substitution Method

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
u  4u 1  0
u  4, u  1
Factor
Set each factor = 0
and solve
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
u  4u 1  0
2
u  4, u  1
Factor
Set each factor = 0
and solve
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
1
2
 4z
1
4
40
u  4u  4  0
2
u  2u  2  0
1
4
uz 2
(z )  z
4
2
2
So let u = z1/4 and get u2 = z1/2.
Substitute u and u2 for z1/4 and z1/2.
Factor & set each factor
u2
= 0 and solve
Solve for z by raising both
1
sides to the 4th power
4
4 4
( z )  (2)
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
x  7x  8  0
6
6
u  7u  8  0
2
u  8u 1  0
u  8, u  1
3
So let u = x3 and get u2 = x6.
Substitute u and u2 for x3 and x6.
Factor & set each factor = 0 and solve
x  8, x  1
3
3
Solve for x by taking the cube root of both sides
x  2, x  1
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au