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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:

Answer

---

2. Solve

Answer

 

Need a break? Play a math game. Well, they all involve math... No, really!

Chapter Contents

• Quadratic Equations
• 1. Solving Quadratic Equations by Factoring
• 2. Completing the Square
• 3. The Quadratic Formula
• 4. The Graph of the Quadratic Function
• 5. Equations in Quadratic Form

• Comments, Questions?

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