factoring imaginary numbers
- Thread starter ca.chick
- Start date
You can't factor \(\displaystyle \L x^{2} - 3x - 2\) using the standard method. Your two solutions for x will be irrational values
Pka completed the square to find the two irrational roots.
\(\displaystyle \L x^{2} - 3x = 2\)
\(\displaystyle \L x^{2} - 3x +(\frac{1}{2}b)^{2}= 2+(\frac{1}{2}b)^{2}\)
\(\displaystyle \L x^{2} - 3x + \frac{9}{4} = 2 + \frac{9}{4}\)
\(\displaystyle \L (x-\frac{3}{2})^{2} = \frac{17}{4}\)
\(\displaystyle \L x = \frac{3}{2}\) \(\displaystyle \small (+/-)\)\(\displaystyle \sqrt{\frac{17}{4}}\)
Pka completed the square to find the two irrational roots.
\(\displaystyle \L x^{2} - 3x = 2\)
\(\displaystyle \L x^{2} - 3x +(\frac{1}{2}b)^{2}= 2+(\frac{1}{2}b)^{2}\)
\(\displaystyle \L x^{2} - 3x + \frac{9}{4} = 2 + \frac{9}{4}\)
\(\displaystyle \L (x-\frac{3}{2})^{2} = \frac{17}{4}\)
\(\displaystyle \L x = \frac{3}{2}\) \(\displaystyle \small (+/-)\)\(\displaystyle \sqrt{\frac{17}{4}}\)
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,990
One of the most fundamental theorems is mathematics is the factor/root theorem. If \(\displaystyle p(x)\) is a polynomial then \(\displaystyle \alpha\) is a root of \(\displaystyle p(x)\) if and only if \(\displaystyle (x-\alpha)\) is a factor of \(\displaystyle p(x)\).axrw said:
That is the whole point of my posting.
- Joined
- Feb 4, 2004
- Messages
- 16,582
Well, as the one tutor displayed, the factored form does not involve imaginary numbers. And when you set the quadratic equal to zero and apply the Quadratic Formula, you will confirm this, when you arrive at irrational, but still real, zeroes.ca.chick said:this problem was meant to have imaginary numbers when you factor it out and im clueless
The point of the other tutor was that this quadratic cannot be factored over the rationals. It can certainly be factored, as the above factorization attests.
If your instructor thought that this quadratic had imaginary zeroes, you might want to print out this thread and ask him to correct or confirm the previous tutors' statements. That way, you're not being confrontational; the "quarrel" comes from us, and you're just "confused" and innocently asking a question. :wink:
Eliz.
ca.chick said:this problem was meant to have imaginary numbers when you factor it out and im clueless
If you're told that something of the form ax^2 + bx + c has two imaginary factors, then check what the expression "b^2 - 4ac" is. If it's negative, it certainly does.