Common in Quadratic Equation when D = 0 or there is a repeated real solution?

Question

For the quadratic equation ax^2 + bx + c = 0,
the discriminant is D = b^2 - 4ac.

1) When D > 0 - There are 2 different real solutions
2) When D = 0 - There is a repeated real solution
3) When D < 0 - There are 2 different complex solutions

There is supposedly a common element(s) in the quadratic equations whose discriminant equals 0 and has a repeated real solution . What is common between all quadratic equations that have a repeated real solution or where D = 0?????

Here are some quadratics with a repeated real solution:

x^2 + (2 square root 2) + 2= 0
D = b^2 - 4ac
D = (2 square root 2)^2 - 2x4x1
D = 8 -8
D = 0 =0
Therefore there is a repeated real solution

x^2 -10x + 25 = 0
D = b^2 - 4ac
D = (-10)^2 -4x25x1
D = 100 - 100
D = 0 = 0
Therefore there is a repeated real solution

SO WHAT IS COMMON BETWEEN ALL THESE QUADRATIC EQUATIONS??????????????

Answer 1

Remember that if x1 and x2 are solutions to the quadratic equation ax²+bx+c=0, then you should be able to divide both sides by a (if a is not already equal to 1) and factor the left into (x - x1)(x - x2) = 0. If there are repeating solutions, then this just becomes (x - x1)² = 0. So I guess one relation these repeating solution quadratics have (besides having a discriminant value of 0) is that they can be factored into a perfect square.

Answer 2

What is common is that they can all be written as
(x - r)² = 0 where r is the repeated root. If you start with the canonical form
ax² + bx + c = 0 and divide through by a you get
x² + px + q = 0. Also p = 2√q and the repeated root is √q

Also, the geometric interpretation is that the parabola described by y = ax² + bx +c touches the x-axis in one and only one point. If D > 0 it touches it in 2 points and, if D < 0, it fails to touch the x-axis at all.


HTH ☺


Doug

Answer 3

The answer is in the text: the solution to the quadratic equation is repeated (i.e. equal). If (b^2 - 4ac) = 0, then x in the quadratic equation would be -b/2a, so you actually have only a single answer.

Answer 4

enable us to be certain the determinant for this that's in many situations B² - 4AC for a quadratic of sort Ax² + Bx + C = 0 here, the determinant is (c-a)² - 4(b-c)(a-b) which equals c² - 2ac + a² - 4{ ba - b² -ca + cb] equals c² - 2ac + a² -4ab + 4b² + 4ac -4cb equals a² + 4b² + c² +2ac -4ab -4cb which could could equivalent 0 for the given quadratic to have a repeated answer which could propose a² + 2ac + c² +4b² -4ab -4cb =0 so (a+c)² = 4b(a+c) - 4b² so (a+c)² + 4b² = 4b(a+c) If a=b=c=0 is actual, then the above is actual If b=0 and the two of a or c isn't 0 then the above isn't actual because of the fact top part is 0 and left part (a+c)² is unavoidably advantageous, till a= -c . if b would not equivalent 0, then b can equivalent a million and c can equivalent a-a million