# Change of Variables

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

The Idea: If you can't solve it **here**, then move **somewhere else** where you 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, you could have solved that without doing that "u=x+1" thing, but here is question where "changing variables" is very useful:

### Example: (x^{2}+2)^{2} - 2(x^{2}+2) - 15 = 0

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

Let u = x^{2}+2, then our equation becomes:

u^{2} - 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 "x^{2}+2":

First Solution | Second Solution |
---|---|

u = 5 | u = -3 |

x^{2}+2 = 5 |
x^{2}+2 = -3 |

x^{2} = 5-2 = 3 |
x^{2} = -3-2 |

x = ±√3 | x^{2} = ±√(-5) |

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

Answer: x = ±√3

Check: ((√3)^{2}+2)^{2} - 2((√3)^{2}+2) - 15 = = 5^{2} - 2·5 - 15 = 25-10-15 = 0

Check: ((-√3)^{2}+2)^{2} - 2((-√3)^{2}+2) - 15 = = 5^{2} - 2·5 - 15 = 25-10-15 = 0

### Example: 3x^{8} + 5x^{4} - 2 = 0

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

But if we use:

u = x^{4}

Then it becomes:

3u^{2} + 5u^{} - 2 = 0

Which **is** Quadratic. And solving it gives:

**u = 1/3** or **u = -2**

Now put the original back again:

First Solution | Second Solution |
---|---|

u = 1/3 | u = -2 |

x^{4} = 1/3 |
x^{4} = -2 |

x = (1/3)^{1/4} |
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 you solve difficult questions.