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A couple months ago, I was challenged to discover a new way to solve quadratic equations with one rule: I must think of it myself. At first I was hopeless, trying various sorts of random things, but I finally found one way. My way is somewhat limited (extremely disappointing), however, but it still works in most cases. You must only know two things about the quadratic equation:

  1. The leading coefficient of x^2 is 1
  2. The difference of the two possible values of x

For example, you have a randomly-picked polynomial with the requirements stated above:

x^2+5x+6=0
Difference:1

Psst! The answer is:

x={-3,-2} 

Which is derived from:

(x+3)(x+2)=0

Obviously, this can be solved in the matter of a second by simple factoring, but there is another way:

let y = greater value of x (-2)
let z = lesser value of x (-3)

Now, you have two new variables, y and z, but how do you use them? Make new equations!

y^2+5y+6=0
z^2+5z+6=0

Somehow, I pulled two magic equations out of thin air:

y^2-(difference*y)=coefficient c
z^2+(difference*z)=coefficient c

So, for our current example, the two equations would look like this:

y^2-(1)y=6
z^2+(1)z=6

Yay! Now we have a system of equations with like variables! The rest can be solved with simple linear combination.

y^2-1y=6 and y^2+5y+6=0 (y^2+5y=-6)

Adding the two equations gives:

2y^2+4y=0

Divide the whole equation by 2:

y^2+2y=0

Subtract 2y from both sides of the equation:

y^2=-2y

Divide both sides by y:

y=-2

Yay! That is the correct answer from above! Now, let's start the same super long process for z(I totally don't want to do this)!

The two equations for z again so you don't have to look up there again(I know, I'm a nice person):

z^2+1z=6 and z^2+5z+6 (z^2+5z=-6)

Adding the two equations together gives:

2z^2+6z=0

Divide the whole equation by 2 gives:

z^2+3z=0

Subtracting 3z from both sides of the equation gives:

z^2=-3z

Dividing both sides of the equation by z gives:

z=-3

Yay! The other solution for x! Now we have the same solution set as we had by factoring:

x={-3,-2}

Phew! Now that my past hour has been used writing this post, I am so tired! You guys are probably really tired too from reading a post from a crazy middle-schooler! Can you please try to help me expand my idea? Help will include:

1. Constructive Criticism
2. Deriving a Formula to Speed up this process
3. Find more ways to use this in quadratic equation solving

Thanks so much!

P.S. I didn't actually have to write out every step. I just wrote it for clarification for the readers.

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  • $\begingroup$ Divide both sides by $y$? (Or by $z$ later on) You cannot do that! $y=0$ is a solution to that particular equation. I think I will stick with the Method of Completing the Square as the general approach for solving quadratic equations... $\endgroup$ – imranfat Nov 24 '15 at 21:57
  • $\begingroup$ @imranfat I know I am degrading the degree of the equation, but y=0 isn't a solution to the original equation. I guess I can write y does not equal to 0, but I didn't think it was quite necessary. Thanks for pointing that out! $\endgroup$ – babyguineapig11 Nov 24 '15 at 21:59
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    $\begingroup$ It seems like this method is just so much longer than the methods already out there...That said, if you are a middle schooler, then certainly the interest you show in math is noteworthy... $\endgroup$ – imranfat Nov 24 '15 at 22:21
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    $\begingroup$ If you know the difference $D= x_1-x_2$ between the two solutions of a quadratic $x^2 + bx + c = 0$ then you can more easily derive the solutions as $x_1 = \frac{D - b}{2}$ and $x_2 = \frac{-D-b}{2}$. This comes from expanding the right hand side of $x^2 + bx + c = (x-x_1)(x-x_2)$ and comparing to get $b=-x_1-x_2$. $\endgroup$ – Winther Nov 24 '15 at 22:38
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    $\begingroup$ this doesn't feel like solving an equation. It feels like having the solutions and then cooking up an equation that satisfies them. $\endgroup$ – Mirko Nov 25 '15 at 0:18

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