Can you use the b^2-4ac thing to show that an equation like x^7 -2x^4 -x^3 +6x +6 has real roots or something. Please explain why not or why it can be used and when it can be used.
Examples provided can be useful.
2006-12-03 16:26:34 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
so can you factor te equation and use this b^2-4ac thing?
2006-12-03 16:33:59 · update #1
lol its hard to decide whos right with two opposing viewpoints
2006-12-03 16:34:49 · update #2
argggggggg
2006-12-03 16:50:25 · update #3
No! The discriminant is only valid for second order.
There is some info available by looking at coefficients. Roots come in complex conjugate pairs, which means that an odd order poly has at least 1 real root.
There are a whole bunch of tricks relating such things. I'd check Wikipedia under polynomials:
You can also make statements about the large |x| behavior. Since 7 is odd and there's a +1 coeffieient it means that p(x)-> -inf as x-> -inf and p(x) -> +inf for large x.
2006-12-03 16:34:05 · answer #1 · answered by modulo_function 7 · 0⤊ 0⤋
Absolutely no, my dear. As a math teacher, i know b^2 - 4ac is only used in quadratic equations. This is called as the discriminant for it tells the nature of the roots of a given quadratic equation. Any equations of higher order or degree cannot be determined by b^2 - 4ac as to the nature of their roots
2006-12-03 23:39:44 · answer #2 · answered by Sheila 2 · 0⤊ 0⤋
b^2-4ac can be used for quadratic equations to determine if they have real roots. if you can factor this equation, then it might be helpful.
every odd degree polynomial has at least 1 real root, since complex roots come in pairs. i'm not sure what question you're trying to answer, but that might help.
2006-12-03 16:30:41 · answer #3 · answered by socialistmath 2 · 0⤊ 0⤋
No, the quadratic formula is not going to be of much help with a 7th order polynomial. It only works for second order ones.
If you could factor the polynomial, then you could find the roots, and then you would know whether they are real.
Alternatively, you could use a computer program to numerically calculate the roots. That's usually easier than factoring.
2006-12-03 16:44:39 · answer #4 · answered by Bill C 4 · 0⤊ 0⤋
No, it's only for quadratic equations of the form
ax^2 + bx + c = 0. That's because if you solve this for x, you get
x = (-b +/- sqrt(b^2 - 4ac))/2a. Equations of degrees greater than two have other solutions.
2006-12-03 16:32:40 · answer #5 · answered by banjuja58 4 · 0⤊ 0⤋