Inactive Tutor answered • 12/10/14
Tutor
New to Wyzant
(5^1/2-3i)^2
To Simplify:
5^1/2-3i(5^1/2-3i)
5-5^1/2*3i-5^1/2*3i+9(i^2). Remember i^2=-1
5 -2(5^1/2*3i)-9
-4-2(5^1/2*3i)
To solve, meaning to rationalize, or remove all irrationals, make = to 0. Then complete the square and sqrt5 and6i will become real numbers.
Mult. By -1
4+2(5^1/2)(3i)=0
4+6(5^1/2)(i) or 4+6i(sqrt5)
Then complete the square to rationalize:
4+6i(sqrt5)=x
4+6i(5^1/2)*(4-6i(5^1/2))=x
16+24i(5^1/2)-24i(5^1/2)-36(i^2)(5) remember i^2=-1
16-36(-1)(5)=x
16+180=196
X=196
See, complete the square on the simplified answer -4-6i(sqrt5)
-4-6i(sqr5)(-4+6i(sqrt5))=x
16-24(6i)(sqrt5)+24(6i)(sqrt5)-36(1)(-5)=x.
The reason for -36(1)(-5) is when you distribute -6 into i^2 and (sqrt5)
You multiply -6*-1 then -1, which is the coefficient in front of -6 or (-1)(6). So (-1)(6)(6)=-36. Then, -1(-1 or i^2)=1. Then when you distribute into (sqrt5^2), you get (-1)(sqrt5^2)or
(-5).
16+ 180=196
X=196
That is why it is easier for me to multiply by -1 before rationalizing. You might miss the -1 coefficient in front of (-6i(sqrt5)) when distributing into the 6i(sqrt5).
Inactive Tutor
Linda,
First remember that the equation = 0. So since you have an equation, you can multiply, divide, add, or subtract any number as long as you do it to both sides of the equation. Most textbooks teach that when you simplify, try to have a positive first term
in your equation. You don't have to, but it makes it easier to work with. So since I still had some simplifying to accomplish, I did this:
-4-2(sqrt5*3i)=0
-1(-4-2(sqrt5*3i=-1(0)0 (multiply both sides by -1)
This becomes 4+2(sqrt5*3i)=0 which also is -4-2(sqrt5*3i). They are equivalent terms.
Then it is easier to go on simplifying with the distribution of 2(sqrt5*3i):
4+2(sqrt5)(6i)
But, it makes no difference. You can leave it as-4-2(sqrt5*3i) if that's as far as your assignment wants you to take because both answers are correct.
Hope that helps..
Thanks
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12/10/14
Inactive Tutor
Also I liked your way of solving. I wanted to add that another way to look at it. Since this is written as standard distribution, it is better to leave it -4-6i(sqrt5). But if they wanted to rationalize, I would go ahead and multiply by-1. But doesn't
look like they wanted to solve for a real number.
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12/10/14
Linda C.
tutor
Thanks for the response. I didn't see this particular wording as an actual equation, but rather just practice combining i terms. I see where you were going with it now.
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12/10/14
Inactive Tutor
You're right not enough is given to know if they want it solved or just simplified. Thanks for pointing that out.
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12/11/14
Linda C.
12/10/14