Question

Find all real and complex roots of the equation below.

x^4 - 81 = 0

Answer 1

a) [-9,9,-9i,9i]
b) [-3,3]
c) [-9,9]
d) [-3,3,-3i,3i]

Answer 2

@Mehek14

Answer 3

Hey Tommy,
remember your difference of squares formula?
\(\large\rm a^2-b^2=(a-b)(a+b)\)
We can apply this rule to 4th powers as well (since 4=2*2)
\(\large\rm a^4-b^4=(a^2-b^2)(a^2+b^2)\)

Answer 4

So we'll start by rewriting 81 as some 4th power, if we're able to.
Can you think of a nice way to break down the 81?

Answer 5

Example: 27=3*3*3
How bout 81?

Answer 6

9?

Answer 7

Well we could certainly write 81 as 9 squared, \(\large\rm 81=9^2\)
But it would be more helpful if we could write it as a fourth power.
So let's rewrite our 9 as 3 squared,
\(\large\rm \color{orangered}{9}^2=(\color{orangered}{3^2})^2=3^4\)
Ok so there is 81 written as a 4th power,
it's 3 times 3 times 3 times 3.

So our expression looks like this,
\(\large\rm x^4-81\)

\(\large\rm x^4-3^4\)

Answer 8

From here we'll apply our difference of squares rule,\[\large\rm x^4-3^4=(x^2-3^2)(x^2+3^2)\]

Answer 9

This at least gets us closer to our roots.

Answer 10

Notice what we end up with is another set of `difference of squares`.
x^2 minus 3^2.

Use the formula again.

Answer 11

Confused? :O
What do you think?

Answer 12

x-3

Answer 13

Sure, that'll be one of the factors,
we should get an x+3 as well though, ya?

Answer 14

\(\large\rm \color{orangered}{(x^2-3^2)}(x^2+3^2)\)
Applying the rule,
\(\large\rm \color{orangered}{(x-3)(x+3)}(x^2+3^2)\)

Answer 15

We're looking for roots, so we'll set our expression equal to 0,
\(\large\rm (x-3)(x+3)(x^2+9)=0\)
and then apply your `Zero Factor Property`:
\(\large\rm (x-3)=0\qquad\qquad(x+3)=0\qquad\qquad (x^2+9)=0\)
and solve for x in each case.
The last equation should give you 2 roots, since it's a squared x.

Answer 16

so would it be b?

Answer 17

Maybe :)
Why b?

I don't like guessers lol :D

Answer 18

cause wouldnt they be 3,-3

Answer 19

Ok good, 3 from the first factor,
-3 from the second factor,

what about that last equation?
\(\large\rm x^2+9=0\)
does this give us any roots?

Answer 20

yes?

Answer 21

Maybe, solve for x :)
Don't be lazy lol :D

Answer 22

3

Answer 23

Subtract 9 from each side,
\(\large\rm x^2=-9\)
Square root each side,
\(\large\rm x=\pm\sqrt{-9}\)
From there, recall what happens when you take the root of a negative number.

Answer 24

so it is b?