Math Question, nothing to work out?

Question

How do you know if a quadratic equation will have one, two, or
no solutions?
How do you find a quadratic equation if you are only given the solution?
Is it possible to have different quadratic equations with the same solution?

Answer 1

To know if there will be one, two, or no solutions, the key is in calculating b^2 - 4ac, also known as the discriminant.

If b^2 - 4ac is a positive number, then there will be two solutions.
If b^2 - 4ac is zero, then there will be one solution.
If b^2 - 4ac is a negative number, then there will be no real solutions.

Example:
x^2 + 8x + 4 will have 2 solutions, since 8^2 - 4(1)(4) = 48.
x^2 + 4x + 4 will have 1 solution, since 4^2 - 4(1)(4) = 0.
x^2 + x + 4 will have no real solutions, since 1^2 - 4(1)(4) = -15.

Answer 2

Let ax^2+bx+c be the quadratic equation
Then if b^2-4ac < 0 , no real solutions
if b^2-4ac = 0, then 1 solution
If b^2 -4ac>0 then 2 different roots.

Iyou are given the roots say x =1 and x=2, then the quadratic equation will be y = (x-1)(x-2) = x^2 -3x +2


Yes. A parabola can be determined by three noncolinear points. So if two of the points are the roots of the equation, yo can select the 3rd point wherever you like and thus have an infinite number of parabolas with the same roots.

Answer 3

Ax^2 + Bx + C = 0

Look at the discriminant, the thing under the square root sign.

That's B^2 - 4AC

If it is positive, you'll have two distinct real roots. If it is zero, you will have one double real root. If it is negative, you will have two complex distinct roots.

If you're told that x = x1 and x = x2, you can make your own quadratic like this (x - x1)(x - x2) = 0 and expand it out.

Yes, it is possible to have different values for A,B,and C and get the same answers. They could differ by a multiplicative constant and result in the same answer.

Answer 4

x^2+bx+c=0
no (real) solutions if b^2 < 4c
one solution (sqrt(c)) if b =+ -2sqrt(c)
otherwise 2 solutions.

suppose x=d, x=e are solutions to quadratic
quadratic is (x-d)(x-e) = x^2 -(d+e)x + de

no. except for constant multiplier
x^2+3x+2=0 has same solutions as 5x^2+15x+10=0 (-1, -2)