Is it true that every quadratic equation can be worked out by either using the quadratic formula or using factorisation?
2006-08-02 22:49:56 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
yes. the formula always works. Then the factors are (x-root1) and (x-root2).... or + lol:)
2006-08-02 23:41:22 · answer #1 · answered by blind_chameleon 5 · 0⤊ 0⤋
You can use the quadratic formula for any quadratic, whether it has real or complex roots. Not all quadratic equations can be solved using factorization because many exist that cannot be factorized.
An example of a non-reducible quadratic:
x^2+x+1=0
x = [-1+sqrt(-3)]/2 or x = [-1-sqrt(-3)]/2
=> x = [-1+i*sqrt(3)]/2 or x = [-1-i*sqrt(3)]/2
=> x = -1/2+i*(sqrt(3)/2) or
x = -1/2-i*(sqrt(3)/2)
Check:
(x+1/2-i*(sqrt(3)/2))
(x+1/2+i*(sqrt(3)/2))=0
x^2 + x/2 + i*(sqrt(3)/2)*x + x/2 + 1/4
+ 1/2*i*(sqrt(3)/2)) - i*(sqrt(3)/2)*x
- 1/2*i*(sqrt(3)/2) + 3/4 = 0
x^2 + x + 1 = 0
2006-08-03 10:32:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋
There is a form for complex numbers, though I don't recall the specifics. Just use i = SRT (-1). i^2 = -1. Seems like a weird thing for Gauss to consider, but who could question his genius. Oh well, since a quadratic is defined as having a maximum term of x^2 (no conics here), is yes, as long as you define i.
2006-08-03 06:08:36 · answer #3 · answered by TurboChemist 1 · 0⤊ 0⤋
I'm not sure about factorization, but using the quadratic equation you can't go wrong.
2006-08-03 05:54:05 · answer #4 · answered by Philbert 3 · 0⤊ 0⤋
Definitely. If there are real roots, factorization is a way to go. Quadratic equation will give you the real roots too, but it will also give you the complex roots! Good times!!! :-)
2006-08-03 08:53:23 · answer #5 · answered by Chris 2 · 0⤊ 0⤋
yes, quadratic formula always gives you the right answer. Factoring may sometimes cannot be use because there are polynomials that you can't factor.
2006-08-03 08:55:00 · answer #6 · answered by xavierbondoc_15 1 · 0⤊ 0⤋
yup, definitely using quadratic formula
2006-08-03 05:58:16 · answer #7 · answered by edmund372 2 · 0⤊ 0⤋
Yes because
ax^2+bx+c = 0
=>x^2+b/ax+c/a =0
=>x^2+b/a+(b/2a)^2-(b/2a)^2+c =0
=>(x-b/2a)^2 = -c+b^2/4a =(b^2-4ac)/2a
take sqrt of both sides +/-
we get x.
2006-08-04 23:22:41 · answer #8 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
after the 7th degree there isnt a formula for polynomials.
2006-08-03 05:54:11 · answer #9 · answered by Kyle M 6 · 0⤊ 0⤋
Not sure if you can deal with imaginary numbers.
2006-08-03 05:55:01 · answer #10 · answered by Anonymous · 0⤊ 0⤋