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: (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 | Second Solution |
|---|---|
| u = 5 | u = -3 |
| x2+2 = 5 | x2+2 = -3 |
| x2 = 5-2 = 3 | x2 = -3-2 |
| x = ±√3 | x2 = ±√(-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 = = 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 | Second Solution |
|---|---|
| u = 1/3 | u = -2 |
| x4 = 1/3 | x4 = -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.