Question

Solve for the roots of x in each of the equations below. a. x4 - 81 = 0 b. x4 + 10x2 + 25 = 0 c. x4 - x2 - 6 = 0

Answer 1

A. x= 3 x=-3

B. No solution

C. x=
`√(3)`

Explanation:

To answer the first one you just have to factorize into conjugate binomials.

`x^(4)-81`=
`(x^(2) +9)`
`(x^(2) -9)`

(x-3)(x+3)
`(x^(2) +9)`

Now you put the binomials in the parenthesis and equal them to 0.

x-3=0

x=3

x+3=0

x=-3

`(x^(2) +9)`=0

`x^(2)`=-9

x=
`√(-9)`

Since the square root of a negative number aint possible the only solutions for the equation are x=3 and x=-3

For the second one you just factorize it into two factors:

`x^(4)+20x^(2)+25`=
`(x^(2) +5)(x^(2) +5)`

Since both solutions end up in x=
`√(-5)` is an imaginary number and it has no solution.

The last one is also factorized into two binomials.

`x^(4)-x^(2)-6`=
`(x^(2) -3)(x^(2) +2)`

Now we put the content of the parenthesis into an equation.

`x^(2) +2=0\\x^(2) -3=0`

Since the +2 will end up ina negative square root it is not possible.

So the only answer for this is x=
`√(3)`

Answer 2

a) 0 = x⁴ - 81

0 = (x² - 9)(x² + 9)

0 = (x - 3)(x + 3)(x² + 9)

0 = x - 3 , 0 = x + 3 , 0 = x² + 9

x = 3 , x = -3 , x² = -9 → x = √-9 → x = +/- 3i

Answer: real roots 3, 3 ; complex roots 3i, -3i

b) 0 = x⁴ + 10x² + 25

0 = (x² + 5)²

0 = x² + 5

x² = -5 → x = +/-√-5 → x = +/- i√5

Answer: no real roots ; complex roots i√5, -i√5

c) 0 = x⁴ - x² - 6

0 = (x² - 3)(x² + 2)

0 = x² - 3 0 = x² + 2

x² = 3 → x = +/-√3 x² = -2 → x = +/-√-2 → x = +/-i√2

Answer: real roots √3, -√3; complex roots i√2, -i√2