In quadratic equations, does the formula work *every* time?

Answer 1

Yes, the result given by the formula always satisfies the equation, though the result may be imaginary.

Answer 2

Yes the formula works every time foor quadratic equations.

Answer 3

Yes, it is possible to obtain roots for every single qudratic equation. When using the formula of x=(-b+-sqrt(b^2-4ac))/(2a), you may get the square root of a negative which is then solved using complex numbers (imaginary numbers that people invented so that we could calculate negative square roots).

In complex numbers, the square root of (-1) is equal to "i"
So an example of a complex number application would be:
2i*2i=(2i)^2
=4(i)^2
=4*sqrt(-1)*sqrt(-1)
=-4

Answer 4

Yes. You should know how to derive this formula.You start off with ax^2+bx+c=0 and then solve for x.

First divide throughout by a:

x^2+(b/a)x+c/a = 0

Complete the square:

x^2+(b/a)x+(1/4)*[(b^2)]/(a^2)]-(1/4)*[(b^2)]/(a^2)]+c/a=0

[x+(1/2)(b/a)]^2-(1/4)*[(b^2)]/(a^2)]+c/a=0

Rearrange terms:

[x+(1/2)(b/a)]^2=(1/4)*[(b^2)]/(a^2)]-c/a

[x+(1/2)(b/a)]^2=(b^2-4ac)/(4*a^2)

Take square root of both sides:

x+(1/2)(b/a)=sqrt[b^2-4ac]/(2a)

x=-(1/2)(b/a)+sqrt[b^2-4ac]/(2a)

This is the final formula:
x={-b+sqrt[b^2-4ac]}/(2a)

Answer 5

Works every time as long as you use it correctly!

Answer 6

yes,every time.even when the roots are imaginery or irrational