Math Is Fun Forum

Kiran

Member

Registered: 2006-11-15

Posts: 177

correct the way i done it

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?

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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: correct the way i done it

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 )

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: correct the way i done it

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.

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