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# FP1

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1. hey guys, i have some difficulties when doing FP1, which says
Find the square roots of 5-12i
The following is my working

finally i get the answer but i have one question
i. why the final answer is not ±(3 +2i)?
by the way, the answer is ±(3 − 2i)
2. You're not far off the full solution, though you made a bit of a mess when solving the quartic.

Firstly, a,b are real.

So when you had your solutions for b, you can discard the two complex ones, and are just left with b=2 or -2.

Then using a= -6/b, you can work out the "a" value in each case, and you then have the two desired solutions. Done.

When solving the quartic, you should recognize that this is a quadratic in b^2. You could use a substitution of c=b^2, though it's not necessary.

So factorising you get (b^2-4)(b^2+9) = 0

Since b is real, b^2 is positive, and so the only solution is b^2=4, and hence b=+/- 2....

Edit: In response to your last point.

If b=2, then a=-3, so one solution is -3+2i
If b=-2, then a=3, so other solution is 3-2i
Hence +/- (3-2i)
3. You're too fast, ghostwalker ^^

Anyways, another factoring tip you might like to know is that when you have a cubic with four terms, such as

I always try to factor by grouping first. For example, if you realize that 1-to-2 is the same proportion as 9-to-18, then you don't have to do synthetic or long division. You can just identify the common factor of b+2 and factor it out.

Note: Another way to interpret this step is that you're taking out the common factors in each group and conveniently being left with two identical remains.

4. (Original post by aznkid66)
You're too fast, ghostwalker ^^
Wasn't quick enough - OP made a significant edit whilst I was working on a reply.
5. IIRC you always get 4 roots of a complex number for sone reason, so my answers would always be
+/- a +/- bi.

Never could get the hang of pulling out the +/-.

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6. (Original post by Snakefingers13)
IIRC you always get 4 roots of a complex number for sone reason, so my answers would always be
+/- a +/- bi.

Never could get the hang of pulling out the +/-.

This was posted from The Student Room's iPhone/iPad App
Hm? When the value of 'a' is uniquely dependent (one-to-one correspondence) on the value of 'b', than the number of valid combinations is the number of solutions for 'b'.

For example, if you had "I have 4 more cats than dogs, and I have ±2 dogs," then 2 dogs 6 cats would be a solution, as well as -2 dogs 2 cats. However, 2 dogs 2 cats and -2 dogs 6 cats are not solutions.

As a rule, there are always two unique square roots (including complex) for any number, including complex numbers.
For example, the unique roots of i=cis(pi/2) are cis(pi/4) and cis(5pi/4).
7. (Original post by Snakefingers13)
IIRC you always get 4 roots of a complex number for sone reason
Well, I'll have you know the Fundamental theorem of algebra disagrees.
8. Fair enough. My teacher sure screwed that one up.

This was posted from The Student Room's iPhone/iPad App
9. (Original post by aznkid66)
You're too fast, ghostwalker ^^

Anyways, another factoring tip you might like to know is that when you have a cubic with four terms, such as

I always try to factor by grouping first. For example, if you realize that 1-to-2 is the same proportion as 9-to-18, then you don't have to do synthetic or long division. You can just identify the common factor of b+2 and factor it out.

Note: Another way to interpret this step is that you're taking out the common factors in each group and conveniently being left with two identical remains.

Nice post.

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