# Change of Variables

Sometimes "changing a variable" can help us solve an equation.

The Idea: If we can't solve it here, then move somewhere else where we can solve it, and then move back to the original position.

Like this: These are the steps:

• Replace an expression (like "2x-3") with a variable (like "u")
• Solve,
• Then put the expression (like "2x-3") back into the solution (where "u" is).

## Example

Here is a simple example: solving (x+1)2 − 4 = 0.

Replace "x+1" with "u" ... Solve ... Replace "u" with "x+1": ## More Examples

OK, we could have solved that without doing that "u=x+1" thing, but here is question where "changing variables" is very useful:

### Example: (x2+2)2 − 2(x2+2) − 15 = 0

It could be hard to solve, but let's try a change of variables:

Let u = x2+2, then our equation becomes:

u2 − 2u − 15 = 0

Which is a quadratic equation that factors nicely into:

(u−5)(u+3)

And the solutions are simply:

u = 5 or u = −3

But wait! We still need to turn "u" back into "x2+2":

First Solution
u = 5
x2+2 = 5
x2 = 5−2 = 3
x = ±√3

Second Solution
u = −3
x2+2 = −3
x2 = −3−2
x2 = ±√(−5)

The second solution is imaginary (it has the square root of a negative number), so let us just use the First Solution:

Check: ((√3)2+2)2 − 2((√3)2+2) − 15 = = 52 − 2·5 − 15 = 25−10−15 = 0
Check: ((−√3)2+2)2 − 2((−√3)2+2) − 15 = = 52 − 2·5 − 15 = 25−10−15 = 0

### Example: 3x8 + 5x4 − 2 = 0

It sort of looks Quadratic, but it is degree 8 which could be impossible to solve.

But if we use:

u = x4

Then it becomes:

3u2 + 5u − 2 = 0

Which is Quadratic. And solving it gives:

u = 1/3 or u = −2

Now put the original back again:

First Solution
u = 1/3
x4 = 1/3
x = (1/3)1/4

Second Solution
u = −2
x4 = −2
x = (−2)1/4

Answer: x = (1/3)1/4 and x = (−2)1/4

Check: You can check this answer!

## Conclusion

"Change of Variable" can help us solve difficult questions, using the steps:

• Replace an expression with a variable (like "u")
• Solve,
• Put the expression back into the solution (where "u" is)