When you use the quadratic formula can b = 0? For example, could I use it for x^2 -3?

Answer 1

yes b can be zero
only case where it fails is that in -4ac should be a positive
in order to avoid imaginary roots.

eg..
x^2 -3
a=1b=0c=-3
roots are

0+-sqrt(0-4(1)(-3)/2*1
=+-sqrt(12)/2*1
=sqrt(3)

lets take the other way

x^2+3
for this the roots will be imaginary
since -4ac=-4*1*4=-12.

so the roots will be sqrt(3)i

Answer 2

Yes. But then it just reduces to sqrt (-c/a). For example, in your example that would be sqrt (--3/1) = sqrt(3).

Since it's obvious more directly that the solutions to ax^2+c = 0 are +/- sqrt (-c/a), I'm not sure the quadratic formula helps you out much.

Answer 3

Absolutely! In the case of x^2-3,
x={0+/-rt(0^2-4(1)(-3)}/2
=+/-rt(12)/2
=+/-2rt(3)/2
=+/- rt3

Answer 4

technicly you can...
in this case X^2-3=0
then x1,2= [-0 +/- sqrt(0^2-4*{-3})]/2
and so:
x1,2= [0 +/- sqrt(12)]/2
x1 = sqrt(3)
x2 = -sqrt(3)

and that was the proof

how ever , its wast of time , in this form of questions you can just put the 3 in other side and then get it in form of :

x^2=3
x1,2=+/- sqrt(3)

same answer , easier way :)

there are ofcourse other technics...

Answer 5

ya b can be any number either negative, Zero or positive

ax^2 + bx + c = 0

D= square root of b^2 - 4ac

and if the number in square root is negative then we have to introduce an imaginary number "i" called IOTA

e.g. square root of -4 will be 2i

Answer 6

yes