2006-09-06 22:58:26 · 8 answers · asked by Terry B 1 in Science & Mathematics ➔ Mathematics
To solve quadratic equations of the form:
ax^2 + bx + c = 0
one can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
2006-09-06 23:06:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
General form of Quadratic equations is Ax^2 + Bx + C =0
Formula to solve these kind of equations is
x = (-B ± (sqrt(B^2 - 4AC )) / 2A
2006-09-07 14:30:39 · answer #2 · answered by king2006 2 · 0⤊ 0⤋
If u want to solve the quadratic equation : ax^2+bx+c=0
then the roots are: (-b+rootover(b^2-4ac))/2a
and (-b-rootover(b^2-4ac))/2a
The formula is Sridhar Acharya's formula.
2006-09-07 06:45:00 · answer #3 · answered by Innocence Redefined 5 · 0⤊ 0⤋
Given ax^2 + bx + c = 0,
The quadratic formula is:
(-b plus minus square root of (b^2 -4ac)) / 2a
It's kinda hard to write without access to proper mathematical notation, but I hope that helps.
2006-09-07 06:03:08 · answer #4 · answered by Bramblyspam 7 · 0⤊ 0⤋
The quadratic formula
x = (-b ± sqrt(b^2 - 4ac))/(2a)
2006-09-07 14:29:44 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
If your equation is in the form
ax^2+bx+c==0,
then the solution will be
x=(-b+/-sqrt(b^2-4ac))/(2a).
2006-09-07 06:34:52 · answer #6 · answered by Curious 2 · 0⤊ 0⤋
Assume the equation to be ax^2+bx+c=0
first divide both the sides by a so that
x^2+b/a x + c/a = 0
Now add and subtract (b/2a)^2 on L.H.S.
x^2 + b/a x + c/a + (b/2a)^2 - (b/2a)^2 = 0
[x^2+2*(x)*(b/2a) + (b/2a)^2] + c/a - (b/2a)^2=0
[x+(b/2a)]^2 + c/a - (b^2)/ (4a^2)=0
[x+(b/2a)]^2 = -c/a + (b^2)/ (4a^2)
[x+(b/2a)]^2 = (b^2-4ac)/(4a^2)
[x+(b/2a)] = +- sq. root [(b^2-4ac)/(4a^2)]
[x+(b/2a)] = +- [sq. root (b^2-4ac)]/ 2a)
x = [-b+- {sq. root (b^2-4ac)}]/ 2a
2006-09-07 07:26:19 · answer #7 · answered by Amit K 2 · 0⤊ 0⤋
A=B+C
2006-09-07 05:59:16 · answer #8 · answered by Anonymous · 0⤊ 0⤋