x ^4 - 81=0?? using quadratic form.?

Question

i just dont have anyone to help me:(

Answer 1

x^4 = 81
x = 3 (Taking only real positive value for x )
Check
3^4 = 3 x 3 x 3 x 3 = 81

Answer 2

Instead of using x^4, you can write y^2 where y = x^2.

So you would have y^2 - 81 = 0 where y = x^2.

For the quadratic formula we need a,b and c.
a = 1
b = 0
c = -81

And then plug it into the formula.

y = ((-(0) ± sqrt(0^2 - 4(1)(-81))) / 2(1)

Lets simplify that a bit and we get a positive and negative solution:
Positive:

y = sqrt(324) / 2
Which =
y = 18 / 2
y = 9

Now plug in x^2 again for y to find x.
x^2 = 9
x = ± 3

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Negative solution:
y = -sqrt(324) / 2
y = -18/2
y = -9
Plug in x^2:
x^2 = -9
x = sqrt(-9)
x = ± 3i

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Note that the degree of x^4 - 81 = 0 is 4 which means that there are FOUR solutions and that is exactly what we got. Two of them were real solutions and the other two were imaginary numbers (i = sqrt(-1))

The solutions are: +3, -3, +3i and -3i.

If you wanted to do this by factoring you could have changed -81 to -(3^4) to have x^4 - (3^4) and worked from there.

Answer 3

why not factor? It is a difference of 2 squares
(x^2+9)(x^2-9)
(x^2+9)(x-3)(x+3)

X^2 + 9=0 gives imaginary roots because there will be square root of -9. Graphically it does not cross the x axis.

Answer 4

x^4-81=0
x^4-3^4=0
(x^2-3^2)(x^2+3^2)=0
1) x^2-9=0 or 2) x^2=--9
X^2=9 not possible
x=+3 or x= --3
okay!!!!!

Answer 5

x^2 = 9, since x^2 cannot be negative.
x = ±3
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Ideas: You don't need to factor it.

Answer 6

(x-3)(x+3)=0

x=3, or x=-3