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vikrantg4 Group TitleBest ResponseYou've already chosen the best response.0
cubic equation omg
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
z = x+iy this is complex number question
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Well first of all you can find out that P(2)=0
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Therefore z2 must be a factor of 2z^3  2z^2  z  6
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
There will be one + ve real root.)
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
why is it that you have fed 2 into the equation, is this a random number
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
When you try to factorize polynomials you can try plugging in numbers to see which one of them give you a zero, it is a trial and error way to factorize easily
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
So if P(2) gives you zero, z2 must be a factor of p(z)
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Then when you divide 2z^3 + 2z^2z6 by z2
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
You get 2z^2 + 2z +3
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Sum of the Roots , product of the roots
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
So we have the following expression (2z^2 +2z +3)(z2)=0
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Now we have found one solutions z = 2
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Other two are given by 2z^2+2z+3=0
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
and the other sol^n will be Complex numbers...
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Gud work...@baddinlol
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
I am going to complete the square
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
polynomials are fun aren't they
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
nop...@baddinlol use quadratic formula
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
2(z^2 + z + 3/2) = 0 z^2 + z + 3/2 = 0 (z + 1/2)^2 + 5/4 = 0
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
(z + 1/2)^2 = 5/4 z + 1/2 = sqrt(5)i /2, sqrt(5)i /2 z = (sqrt(5)i  1)/2, (1sqrt(5)i)/2
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
that will be =+orsqrt b^24(a)(c)/(2a)
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
well done baddinol thanks
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Did you know about putting in numbers to find p(Z) = 0?
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
By the way the numbers you plug in have to be a factor of the constant in the equation, which in your equation is 6::::: 2z^32z^2z6
 2 years ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
i did but i thought there must be a faster way
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
SO i tried 1, 1, 2 and when i got upto 2 i stopped because i found P(2)=0
 2 years ago

baddinlol Group TitleBest ResponseYou've already chosen the best response.1
Are you allowed to use a cas calculator?
 2 years ago
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