5. Equations in Quadratic Form

In this section, we will be meet equations that are actually quadratic in form, but they may not look like it at first glance.

We will use either of the following methods to solve such equations:

 

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:

math expression

We can see from the graph that the solutions are correct.

 

Example 2:

Solve: math expression

Here, if we write u = √x we have:

math expression

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: math expression OK.

But √x = -1 is not possible (√x is always ≥ 0).

We conclude there is only one root: x = 1/16

 

Exercises

1. Solve:math expression


Answer

2. Solve math expression


Answer



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