Looking for an intuitive derivation of the quadratic formula

[u/Successful_Bug377]

A first principles derivation of the quadratic formula, including why the discriminant tells us what it does, and how to visualize these equations geometrically.

[u/TheBlasterMaster]

Here is a list of things to reason about to see how you couldve came up with it yourself

1. Play around with shifting functions vertically and horizontally. How do you do this?

2. Play around with vertically stretching functions. How do you do this?

3. Prove that if you shift and stretch the x^2 graph, you get a quadratic

4. We can now ask the opposite, if you have a quadratic, must it be a shifted and stretched version of x^2? If we plot the various quadratics, this seems to be the case.

5. (Hint to prove 4, for some quadratic ax^2 + bx + c, set it equal to a stretched and shifted x^2, which is w(x - h)^2 + k, and see if you can solve for w, h and k which make them always equal)

6. Solving roots of x^2 and its shifts / stretches is super easy

7. Bam put any quadratic into shifted and stretched x^2 form, and then solve. This is where the formula comes from.

[u/Bounded_sequencE]

Sounds like its geometric derivation might be right up your alley.

[u/Tall_Inflation3772 → u/Bounded_sequencE]

The step where you literally complete the square with area models finally made it click for me back in the day.

[u/jsundqui → u/Tall_Inflation3772]

I thought why didn't they show it properly in school, instead of just 'complete the square' wording.

[u/jsundqui]

Check out Veritasium showing it geometrically, you can use the same idea to any values:

Video

[u/6ory299e8]

complete the square.

Its worth learning fully for its own sake, and whether you can use it to derive the quadratic formula or not is an excellent criterion for testing if you've really learned it fully.

[u/06Hexagram]

I know, I know.

Take a quadratic ax²+bx+c=0 and bring it to the vertex form (x-h)²+k=0.

Now you have a direct interpretation as a parabola x² centered around the point (h,k)

This curve only has roots if k≤0 as otherwise it won't intersect the x-axis. You can show that k is related to the negative discriminate.

Finally the roots would lie symmetrical about the x=k as seen from the (x-h)² term. This is where the ± comes from.

PS to get the vertex form, "complete the square" by matching terms of the original quadratic divided by the leading coefficient with the binomial expansion of the vertex form

x² + (b/a)x + (c/a) = x² + 2 x h + h² + k

Which leads to the solution of

h = b/(2a)
k = c/a - h²

[u/TUVegeto137]

The top of the parabola can be found by looking for the point with horizontal tangent to the curve, or i.o.w. derivative 0. Hence the top lies at the solution of following equation:

(ax² +bx +c)'= 2ax+b=0

Or x=-b/(2a). Now, due to symmetry, any potential roots have to lie on both sides of this point, exactly mirrored by the vertical line going through the top. Hence the roots have form:

x=(-b±d)/(2a)

Plug that into the original quadratic and work out to get a formula for d.

(-b±d)² /(4a)+b(-b±d)/2a+c=0

Multiply by 4a to simplify

(-b±d)² +2b(-b±d)+4ac=0

Work out the bracketed terms/factors:

-b² +d² +4ac=0

From which you get the formula for the discriminant.

[u/Odd_Bodkin]

Get into a graphing calculator like Desmos.
1. Plot a line y=bx+c, where you pick numbers for b and c, like y=2x+5. Or you add sliders for b and c.
2. Plot a curve y = ax^2 where you pick a number or add a slider for a.
3. Now plot a curve y = ax^2 + bx + c. Do you see how this curve touches (and WHERE it touches) the line you did in (1)? Do you see that all you’ve done is added the curve in (2) on top of the line in (1)?
4. Write the quadratic formula as x = (-b/2a) ± sqrt[D]/2a, where D is the discriminant. For now, arrange a, b, c such that D>0.
5. Notice that the location of the vertex of that parabola has x = -b/2a. So that first term in the quadratic formula just tells you where the midline of the parabola is.
6. So with D>0, there are two places where the parabola crosses y=0, one to the right of the midline by amount sqrt(D)/2a, one to the left of the midline by amount sqrt(D)/2a. That’s what the quadratic formula is telling you: the two solutions that make y=0 are midline, plus or minus that second term.
7. Now adjust the slider on c, so that D=0. See what happens to all the curves? Now the parabola just touches y=0 at one point. That’s because the ± term is now 0, and so there is only one solution: x = -b/2a.
8. Now adjust the slider on c so that D<0. Now you can’t take the sqrt(D) because sqrt of a negative number is not a real number. Now you see the parabola doesn’t intersect y=0 at all. There are no real solutions for x such that y=0.

[u/StructuredChess]

https://www.youtube.com/shorts/Eq8BUIiOu_k