5. Equations in Quadratic Form
In this section, we'll come across equations that are in fact quadratic, but they may not look like it at first glance.
We'll use either of the following methods to solve such equations:
- factoring
- quadratic formula:

Example 1:
Solve: x4 - 20x2 + 64 = 0
Answer
Here, if we let u = x2, we can rewrite the equation so it looks like an ordinary quadratic equation:
u2 - 20u + 64 = 0
We now factor to give:
(u - 16)(u - 4) = 0
So the solutions for u are 16 or 4.
So x2 = 16 or x2 = 4.
These give us:
| x = -4 or 4 | x = -2 or 2 |
So the complete set of solutions is: x = -4, -2, 2, 4.
Is it correct?
The sketch shows:

We can see from the graph that the solutions are correct.
Example 2:
Solve: ![]()
Here, if we write u = √x we have:
So u = 1/4; or u = -1.
DANGER! Always think carefully about your answer. You can often get answers which are not true solutions.
√x = ¼ means x = 1/16
Check by substitution:
OK.
But √x = -1 is not possible (√x is always ≥ 0).
We conclude there is only one root: x = 1/16
Exercises
1. Solve:![]()
2. Solve ![]()
Didn't find what you are looking for on this page? Try search:
The IntMath Newsletter
Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!
Algebra Lessons on DVD
Easy to understand algebra lessons on DVD. See samples before you commit.
More info: Algebra videos
Book mark this page
Add this page to Del.icio.us, Furl, Digg, StumbleUpon, Google, whatever...
Need a break? Play a math game. Well, they all involve math... No, really!





