You are not logged in.
Pages: 1
factor x^2 - 3x + 4 over the set of complex numbers.
i did
the quadratic way
by getting
x = -3 +- sqrt(-7) / 2
and then
x = -1.5 +- sqrt(-3.5)
now what?
Live 4 Love
Offline
you're solving the equation, it wants you to factor it, make it into the form (ax + b)(cx + d)
since its x^2 - 3x + 4, -, + you have the form (x - a)(x - b), so its a case of finding a and b
(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 - 3x + 4
so you have
a+b = -3, a = -3 - b
ab = 4,
(-3-b)b = 4
-3b - b^2 - 4 = 0
b^2 + 3b + 4 = 0, (then by quadratic equation)
b = -1.5 ± 0.5√-7
b = -1.5 ± 0.5i√7
you then have a = -3 - b
so you're two possible factorisations are
4.5 + 0.5is7, 1.5 - 0.5is7
1.5 - 0.5is7, 1.5 + 0.5is7
(check they are right, i cant be math )
The Beginning Of All Things To End.
The End Of All Things To Come.
Offline
Everything looks right there to me, luca.
Soha's method for finding the roots would have worked as well. You can use the quadratic equation to find the roots, then put the roots a and b into (x-a) and (x-b), and you have your factorisation.
However, Soha made an error by saying that √-7/2 = √-3.5.
To divide by 2, you first need to make the 2 into square root form, so √4.
You then divide -7 by 4 to get √-1.75.
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1