calculate the roots of the eq x^2+x -1=0. I really dont want the roots but i want the quadratic to be expressed as multiplication of factors like (x- root1) (x-root2)

Question

anonymous · Answer

Thanks for the answer which is quite hard to understand than using the simple $$(-b  \pm \sqrt{b ^{2}-4ac})/2a $$ when the quadratic is a$$ax ^{2}+bx+c=0$$ but my doubt is: is the factor equation always expressed as (x-root1)(x-root2) or they do change the signs like (x-root1)(x+root2) something like that

phi · Answer

If you use the quadratic formula you get two roots, called r1 and r2
then your factors would be
(x-r1)(x-r2)

phi · Answer

See http://www.wolframalpha.com/input/?i=%28x-%28sqrt%285%29-1%29%2F2%29%28x%2B%28sqrt%285%29%2B1%29%2F2%29

asnaseer · Answer

@phi is right. the reason you can also end up with something like (x-a1)(x+a2) is if one of the roots is a negative number.
e.g.,
1. say the roots were 3 and 7, then this would give you (x-3)(x-7) = 0
2. say the roots were -3 and 7, then this would give you (X+3)(x-7) = 0
3. etc...

asnaseer · Answer

another way of writing my example number 2 is:
(x-(-3))(x-7) = 0

jamesj · Answer

If r1 and r2 are roots of ax^2 + bx + c = 0, then it must be the case that

ax^2 + bx + c = a(x - r1)(x-r2) = 0

asnaseer · Answer

yes, I agree with @JamesJ, I was trying to address @chaotic.clone's question about whether you could end up with (x-something)(x+somethingelse).
I don't know if I was clear enough.

anonymous · Answer

@JamesJ, is it true that if the eq is ax^2+bx+c=0 has the roots of form just (x-r1)(x-r2) but not a(x-r1)(x-r2) as you have said.

jamesj · Answer

What I am saying is that as polynomials,

ax^2 + bx + c = a(x - r1)(x-r2).

The a is strictly necessary, even if it is usually that case that a = 1.