Question

Find all roots. Answer as soon as possible. Explain how to do it.

Answer 1

x = ± 3

Explanation:

`x {}^(4) - 5x {}^(2) - 36 = 0 \\ let \: x {}^(2) = y \: (subtitution) \\ (x {}^(2) ) {}^(2) - 5x {}^(2) - 36 = 0 \\ y {}^(2) - 5y - 36 = 0 \\ (y + 4)(y - 9) = 0 \\ y = - 4 \: \: \: \: \: \: \: \: \: \ \: y = 9 \\ (y = x {}^(2) ) \\ x {}^(2) = - 4 \: \: \: \: \: \: \: \: \: \: \: \: \: x {}^(2) = 9`

We can't take the square root of a negative number so x² = -4 is rejected

`x {}^(2) = 9 \\ x = √(9) = \binom{ + }{ - } 3`

Therefore

x = 3

x = -3

Answer 2

2i, -2i, -3, 3

Explanation:

We can complete the square to factor but make x to the power of 2 since the variable x is squared in the function aswell

Factor

(
`x^(2)` + 4)(
`x^(2)` - 9)

Now we have two function we can set equal to 0

Find zeros

`x^(2)` + 4 = 0

`x^(2)` - 9 = 0

`x^(2)` = 9

`x^(2)` = -4

x =
`√(-4)`

x =
`√(9)`

±2i, ±3