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OpenStudy (anonymous):
x^2 – 5x + 10 = 0
OpenStudy (anonymous):
What are your options?
OpenStudy (anonymous):
x^2 – 5x + 10=0
If you forget the +10 bit for a bit and take a look at x^2-5x only what number you need to add to that to get a perfect square. (perfect square looks like (x+1)^2)
OpenStudy (anonymous):
okay.
OpenStudy (anonymous):
(x+y)^2=x^2+2xy+y^2
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OpenStudy (anonymous):
Sorry in this case more important is
(x-y)^2=x^2-2xy+y^2
OpenStudy (anonymous):
x^2-5x This is the x bit in this case, what is should y be?
OpenStudy (anonymous):
From these 3 parts which one is the -5x equivalent with?
x^2
-2xy
+y^2
OpenStudy (anonymous):
x^2
OpenStudy (anonymous):
wrong guess :)
-2xy
Think y only as a number.
-5x=-2xy
y=5/2=2.5
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
So where are we going with this?
OpenStudy (anonymous):
so it would be:
(x – 5/2)^2 + 15/4 = 0
OpenStudy (anonymous):
x^2 + 10 = 5x
OpenStudy (anonymous):
Oh yes you are right!
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OpenStudy (anonymous):
So how can you solve it now?
(x – 5/2)^2 + 15/4 = 0
OpenStudy (anonymous):
X = 1/2 ( 5 – i sqroot 15)
Or x = 1/2 (5 + i sqroot 15)
OpenStudy (anonymous):
thats the final answer i came up with.
OpenStudy (anonymous):
If you allow complex numbers.
If not you can just say no real solutions
OpenStudy (anonymous):
oh okay but is my answers correct?
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OpenStudy (anonymous):
yes
OpenStudy (anonymous):
i just want you to check on my answers:
Solve 3x^2 – 2x – 9 = 0
3x^2 - 2x = 9
x^2 - (2/3)x = 3
(x - 1/3)^2 = 10/9
x = (1 - sqroot10)/3
x = (1 + sqroot 10)/3