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Mathematics 21 Online
OpenStudy (anonymous):

Solve x2 – 5x + 10 = 0 by completing the square.

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

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

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!

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?

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

OpenStudy (anonymous):

yes

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