x^3 -2x^2+13x=0 leave in exact form no decimals approx

- anonymous

x^3 -2x^2+13x=0 leave in exact form no decimals approx

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- schrodinger

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- anonymous

You could write the same equation as:
x(x^2 - 2x + 13) = 0 so one root is x = 0
and the others are the roots of x^2 - 2x + 13 =0 equation which are
(1 - 2sqrt3) i and (1 + 2sqrt3 i)

- anonymous

did you solved this using quadratic formula negative b + or - square root b^2 - 4ac over 2a?

- radar

I believe he did and he simplified the answer.

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## More answers

- anonymous

i tried that i did broke it down i got 2+ or - sqrt-48 over 2 times "a" which is 2 but when you divide the top (2+ or - sqrt-48) i got 24 which breaks down to to 2i sqrt 6 how did he get the 3?

- radar

\[\sqrt{-48}=\sqrt{16\times-3}\]

- radar

\[\sqrt{16}\times \sqrt{-3}\]\[4\sqrt{-3}\]

- radar

does that simplification rings a bell?

- anonymous

yeah i get it now its hard when someone just leaps from one spot to another

- radar

I haven't double checked but nikola usually gets it right. Good luck

- anonymous

Sweets,
http://en.wikipedia.org/wiki/Quadratic_equation
a=1 b=(-2) c=13
b^2 - 4ac = (-2)^2 - 4*1*13 = -48 which is smaller than 0,so roots are complex and can be calculated as :
-b/2a + (sqrt [- (b^2 - 4ac)] /2a)*i
-b/2a - (sqrt [- (b^2 - 4ac)] /2a)*i

- anonymous

yay nikola lol

- radar

And I have to agree with you sometimes simplified looks more complicated lol\[\sqrt{-48}=4\sqrt{-3}\] doesn't look to much more simple Hi

- anonymous

2+(sqrt[-2^2-4(1)(13)]/2(1) = 2+(sqrt[-48])/2 = so do I divide out the 2 from the bottom like so or no lol sorry for all this... 1+(sqrt[-24] then you break down the 24 to 4i sqrt 6

- anonymous

so 1+\[4i \sqrt{6} and
1-4i \sqrt{6}\]

- anonymous

that was suppose to be 2i sqrt6 not the 4 my bad

- anonymous

Sweets,
a=1 b=(-2) c = 13
b^2 - 4ac = (-2)^2 - 4*1*13 = -48 < 0 so;
-b/2a + (sqrt [- (b^2 - 4ac)] /2a)*i
-b = -(-2) = 2
so -b/2a is 2/2*1 =1
sqrt [- [b^2 - 4ac) ] = sqrt [-(-48)] = sqrt48 which is 4sqrt3
(sqrt [- (b^2 - 4ac)] /2a)*i is 4sqrt3 / 2 *1 = 2sqrt3
I hope this helps.

- anonymous

i have got 2+24i and 2-24i as the answer

- anonymous

yes thats correct nikola:)

- anonymous

Sweets,it s -b/2*a not -b/a :)

- radar

\[(-(-2)\pm \sqrt{(-2)^{2}-(4)(1)(13)})/2\]\[(2\pm \sqrt{4-52})/2\]\[(2\pm \sqrt{-48})/2\] can you take it from there?

- anonymous

Oooooo lol i see i see

- anonymous

you miss one thing it messes it all up lol

- anonymous

yeah...thats what math is all about:)

- anonymous

haha gotta love it

- anonymous

so th eanswer probably is 1+24i & 1-24i if i am not wrong:)

- radar

Continuing on.\[(2\pm(\sqrt{16\times-3})/2\]
\[(2\pm(\sqrt{16}\sqrt{-3})/2\]\[(2\pm(4\sqrt{-3})/2\] now do the division\[1\pm2\sqrt{-3}\]

- anonymous

how did u get -3 there? under the root???

- radar

I would leave it there are you could go further and do this. Convert the radical as follows:\[\sqrt{-3}=\sqrt{3 X-1}\]

- radar

The square root of -1 is the imaginary operator i so the final answer becomes:\[1\pm \sqrt{3}i\]

- radar

Did you understand I was trying to show in the radical 3 times a -1 ?
and i left out the 2 so it should be\[1\pm2\sqrt{3}i\]

- anonymous

yeah i seen what you did 4 squared times 3 is 48 but you had to divide the 4 by 2 i get it

- radar

Yeahh!!!Now practice is the key. Notice that they did not want you to really get the value no decimals etc. so that is as far as you need to take it.

- anonymous

lol thank all of you so much. Im not as slow as i seem this was just a total brain fart

- anonymous

Im a visual learner so it helped when you showed :)

- anonymous

you're welcome sweets,good luck with your studies.

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