Question

X
`x^(4) -41x^(2) +400=0`

Answer 1

Values of x are ±4 and ±5

Explanation:

Let y =
`x^(2)`

Which means:
`y^(2) = x^(4)`

Therefore:

A quadratic equation is formed:

`y^(2) - 41y + 400 = 0`

The above quadratic equation can be simplified either by The Double Bubble Method or using the quadratic formula

Double Bubble Method:

`y^(2) - 41y + 400 = 0`

The pair of factors of 400 that add together to give 41 are 25 and 16. Use these two factors to rewritre -41y:

`y^(2) - 25y - 16y + 400 = 0`

Then do Factorization by grouping (i.e group the terms with common factors before factorizing):

`y(y -25) - 16(y - 25) = 0`

`(y - 25)(y - 16) = 0`

The product of two expressions is zero, which means either of the two expressions is equal to Zero:

Either:
`y - 25 = 0`

Substituting
`y = x^(2)`:

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

=
`x^(2) = 25`

=
`\sqrt{x^(2)} = √(25)`

∴ x = ±5

Or:
`y - 16 = 0`

Substituting
`y = x^(2)`

`= x^(2) - 16 = 0`

`= x^(2) = 16`

`= \sqrt{x^(2)} = √(16)`

∴ x = ±4

∴Values of x are ±4 and ±5