Question

i forgot how to do these

Answer 1

im not sure what the question is

Answer 2

like how do i solve it.

Answer 3

the question is write each equation give the roots

Answer 4

Write an equation that has those roots?

Answer 5

yes here i will show u wat i did

Answer 6

For roots a, b, c, the equation is:
(x - a)(x - b)(x - c) = 0

Answer 7

ik i left out a (x+8i)

Answer 8

You're on the right track.

Answer 9

oh so how do i multi. all of them?

Answer 10

Before we get there, I need to show you something.

Answer 11

Let's say a root is a.
The binomial that has the root is x - a
Ok so far?

Answer 12

yep

Answer 13

This is where you need to be careful with algebra.
Let's say the root is \(2 + \sqrt 3\)
Then the binomial that contains that root is:
\(x - (2 + \sqrt 3)\)
Notice the parentheses around the root.

Answer 14

That means the binomial is \(x - 2 - \sqrt 3\)

Answer 15

isn't there 2 binomials ?

Answer 16

If you leave the parentheses out, then you'd get
\(x - 2 + \sqrt{3}\) which would be incorrect.

Answer 17

Yes, you may have many binomials. I'm only pointing out that you need to be careful with the expression x - a, where a is the root. If "a" has involves a sum or a difference, it must be included in parentheses.

Answer 18

Let's do now problem 12 which you posted above.

Answer 19

yes i got it.
do u know how i could multiple them

Answer 20

\(\large [x - (5 - 2\sqrt{3})][x - (5 + 2\sqrt{3})][x - (-1)] = 0\)

Answer 21

\(\large [x - (5 - 2\sqrt{3})][x - (5 + 2\sqrt{3})](x +1) = 0\)
We can start by multiplying the first two binomials together.

Answer 22

yes okay

Answer 23

First, we need to simplify the insides to remove the inner parentheses:
\(\large (x - 5 + 2\sqrt{3}))(x - 5 - 2\sqrt{3})(x +1) = 0\)

Answer 24

okay here i will try to work it out them i'll send it to u

Answer 25

One way to do it is to multiply every term of the first factor by every term of the second factor. There is one shortcut we can use.

Answer 26

oh okay lets use a short cut because idk how to do it

Answer 27

Are you familiar with the product of a sum and a difference which equals a difference of squares?
For example: \((a + b)(a - b) = a^2 - b^2\)

Answer 28

i think so

Answer 29

Ok. Look at the two first factors. They can be thought of a product of a sum and a difference. I'll use colors to show you that.

Answer 30

\(\large (\color{red}{x - 5} + \color{blue}{2\sqrt{3}})(\color{red}{x - 5} - \color{blue}{2\sqrt{3}})(x +1) = 0\)

Answer 31

do i square x-5 ?