Solve for x. x^4 - 81 = 0?

Answer 1

x^4 = 81 ==> then do fourth root of each side
x = +/- fourthroot of 81 (sqrt of the sqrt)
x = +/- 3

x=3 or x= -3
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OR, if you are more comfortable with factoring, it is the difference of perfect squares which then factors again!
x^4 - 81 = 0
(x^2 - 9)(x^2 + 9) = 0
(x-3)(x+3)(x^2 + 9) = 0
x-3=0 OR x+3 =0 OR x^2+9 = 0
x=3 OR x= -3 (then x^2 = -9 has no real solutions)
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If you need complex solutions (nonreal answers, then there are two more solutions from x^2= -9 ===> x = 3i and x = - 3i
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Answer 2

X^4-81=0
X^4=81
x=3
proof
3^4=81-81=0

Answer 3

You can get a direct answer without factoring at all.

x^4 - 81 = 0
x^4 = 81
x = +/- fourth root of 81
x = 3, -3

Factorisation method: (a^2 - b^2) form
x^4 - 81 = 0
(x^2)^2 - (9)^2 = 0
(x^2 + 9)(x^2 - 9) = 0

Either x^2 + 9 = 0 (or) x^2 - 9 = 0

When x^2 + 9 = 0
x^2 = -9
x = +/-(sqrt - 9)
x = 3i , -3i

(sqrt - 9) is not a real number. It is a complex number. There are no real roots for x^2 + 9 = 0.

When, x^2 - 9 = 0
x^2 = 9
x = +/- sqrt 9
x = 3, -3

which is the same answer we got in a concise manner.
3 and -3 are the real roots of the given equation. 3i and -3i are the complex roots.

i = sqrt (-1)

If you have doubts as to how (sqrt -9) = 3i, - 3i. Here's proof:

sqrt (-9) = (sqrt 9) * sqrt (-1)
= (sqrt 9) * i
= +/-3 * i
= +/- 3i

Answer 4

x^4 - 81 = 0
x^4 = 81

Take square roots:
x^2 = ±9
x^2 = 9, x^2 = -9

For x^2 = 9, take square roots:
x = ±3

For x^2 = -9, take square roots
x = ±3i

So basically we have the four solutions x = ±3, x = ±3i


Another method is factorising:
x^4 - 81 = 0
Difference of perfect squares:
(x^2 + 9) (x^2 - 9) = 0
(x^2 - (-9)) (x^2 - 9) = 0
More difference of perfect squares:
((x + 3i) (x - 3i)) ((x + 3) (x - 3)) = 0
x = -3i, x = 3i, x = -3, x = 3

Answer 5

so thats x^5 - 81 = 0
x^5 = 81
x= 5 th root of 81.
If there's a bracket containing x^4 and 81 in it, the answer is
+3 or -3

Answer 6

x^4 - 81 = 0
(x^2-9)(x^2+9)=0
x^2+9=0X
or
x^2-9=0
(x+3)(x-3)=0
x=-3 or x=+3

Answer 7

x^4-81=0
x^4=81^(1/2)
x^2=9^(1/2)
x=3

Answer 8

It's a fouth order polynomial with real coefficients. The Fundamental Theorem of Algebra says that there are 4 solutions.

In addition to the two real solutions already given, there are also the two complex solutions:

x = 3i
and
x = -3i

Answer 9

x^4 - 81 = 0
(x^2 + 9)(x^2-9) = 0
(x^2 + 9)(x + 9)(x - 9) = 0
x^2 = -9; x = -9; x = 9
x = {i square root of 9, -9 and 9}

Answer 10

(x^2 - 9)(x^2 + 9) = 0
x^2 - 9 = 0
so x = +/- 3

x^2 + 9 = 0
x = +/- 3i