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:

Change of Variable

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":

Change of Variable

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 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 us solve difficult questions.