If i = square root of -1, simplify (2-3i)(4=2i)...can anyone help me with this?

Question

Can anyone show steps for it..or explain it to me..thanks!

Sorry..it should be 4-2i*

Sorry..it should be 4-2i*

Answer 1

I'm confused by the (4=2i) part of the equation. Should that read as (4-2i) instead?

To all others who question that a negative number can be a square root, it is possible, but not with "real" numbers. The square root of -9, for example, is defined as 3i.

Anyway...if you meant (2-3i)*(4-2i).....

(2-3i)
(4-2i)

Multipling bottom to top/left to right, you get...

8 - 12i - 4i + 6i^2

This simplifies to...

8 - 16i + 6i^2

6i equals the square root of -36. So, the 6i^2 part of the equation can be rewritten as...

((-36)^0.5)^2, which is the same as -36^1...or just plain -36

So, now the equation reads as...

8 - 16i - 36...or further simplified as...

-16i - 28...or

-4(4i + 7)

Answer 2

Did you mean (2-3i)(4-2i)? I'm not sure if this is the right solution to the problem.
But if so, use the FOIL method, Multiple the First variables in the parenthesis, then multiply the Outter, then Inner and Last variables. Which gives...

Let i = x , since you cant really find a real answer because i is a negative number within a sq. root. It makes it a bit easier if you are used to x variables.
(2)(4)=8 , (2)(2x)=4x , (-3x)(4)=-12x , (-3x)(2x)= -6x^2

8+ 4x -12x - 6x^2
combine like terms

8 - 8x - 6x^2
you can factor out a 2 for further simplifcation. I'm not sure whether you can factor the i, my algebra is a bit rusty. my solution may not even be correct just to remind you

**marcy_ddd is completely wrong, she uses a distorted version of FOIL, and Imago left out the exponent in the 6

Answer 3

Well there is no square root of a negative number, because any number squared is a positive, try it.

but I guess you could

(2-(3sqrt-1))(4-(2sqrt-1))

take it as a binumial equation.

8-(4sqrt-1)-(12sqrt-1)+6(-1) < (because a sqrt multiplied by a sqrt is the whole number of the sqrt)

which would give you:

8-(16sqrt-1)-6

2-(16sqrt-1)

2-16i

2(1-8i)

I think that's right

Answer 4

It seems to me its impossible because -1 has no square root what so ever. Try it on a calculator and see. Its impossible.

Answer 5

a million. sqrt(6x^2) = sqrt(6)sqrt(x^2) = x sqrt(6). 2. sqrt(27) - sqrt(3) = sqrt(9)sqrt(3) - sqrt(3) Factorise out the sqrt(3): sqrt(3)(sqrt(9) - a million) = sqrt(3)(3 - a million) = 2sqrt(3). No, sorry. you're incorrect.

Answer 6

(2 - (3 * i)) * (4 - (2 * i)) = 2 - 16 i