Using b2-4ac...can you find then number of possible triangles there are?...and how?

Question

PLEASE ANSWER ME I HAVE MATH B TOMORROW!!!!!!!!!!!!!!!!!!

Answer 1

The general form of a quadratic equation in x is:

ax^2 + bx + c = 0
where a is not 0 and a, b, c are real numbers.

b^2 - 4ac = D is called the discriminant of a quadratic equation.

If D > 0, the equation has real and unequal roots
If D = 0, the equation has real and equal roots
If D < 0, the equation has no real roots
If D is a perfect square, the roots are rational
If D is not a perfect square, and D > 0, the roots are real and irrational.


This is the quadratic formula:

x = [-b +/- (sqrt D)]/2a

If any quadratic equation has an irrational root in a + sqrt b form where a is rational and b is not a perfect square, then the other root is necessarily a - sqrt b

Remember,
D = b^2 - 4ac

I don't see a relation between D and triangles

I'm sorry your triangles question was not answered, but you know quite a lot about quadratic equations and the nature of roots now.

Answer 2

b² - 4ac is the discriminant part of the quadratic formula. It doesn't usually have anything to do with triangles. This question needs more of a context.