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embizze eNotes educator| Certified Educator

Solve `x^4=25` :

Write in standard form:

`x^4-25=0`

This is a quadratic in `x^2` :

`(x^2)^2-25=0`

This is a difference of two squares:

`(x^2+5)(x^2-5)=0`

Applying the zero-product property we get:

`x^2-5=0 => x=+-sqrt(5)`

`x^2+5=0 => x^2=-5 => x=+-isqrt(5)`

Thus the four solutions are `+-sqrt(5),+-isqrt(5)` .**If you are asked only for real solutions, the answer given above is correct**

hala718 eNotes educator| Certified Educator

`x^4 = 25 `

To solve, we will take the forth root of both sides.

`==gt ^4sqrt x^4 = ^4sqrt 25 `

Now we will simplify and take absolute value for x.

`==gt I x I = sqrt5 `

`==gt x = +-sqrt5`

Then, there are two 4th roots for 25 : `-sqrt5, and sqrt5`

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