# Someone has rediscovered a simplified way to solve the quadratic equation

I always disliked the idea of having to memorize the quadratic formula.

And, completing the square – the other main approach to finding the roots of a quadratic equation – was also non-intuitive for me. I wanted to relate the solution to the graph of the equation, so neither the quadratic formula nor completing the square worked well for me.

Now, someone has rediscovered a nice visual way to find the roots of a quadratic equation. (For those of you who may be a bit rusty, the roots are where the curve intersects with the X-axis. )

In the case of a quadratic, if the curve intersects the X-axis at two places, then these will be two roots.

y = x^2  +  bx   +  c

If we let x = 0, we can see that c is where the curve intersects the y-axis.

y =  (0)^2  +   (0)x   + c
y = c

Also, from the graph we can see that u is the average of x1 and x2.

u = (x1 + x2) / 2

These two values (i.e. x1 and x2) are the two places the curve intersects the x-axis. They are the roots of the quadratic equation.

Now, for the time being, we’re going to just assume that these two statements are also true.

c =  (x1) (x2)
u = -b/2

From the graph of the quadratic above, we can see that z is the distance between x1 and u. It is also the distance between x2 and u.

Therefore:
x1 = (u – z)
x2 = (u + z)

And, since we’ve assumed that c = (x1)(x2) we can see that:
c = (u – z)(u + z)

If we multiply together the two items in parentheses on the right, we get:

c = u^2 + zu – zu – z^2

Since zu – zu = 0 we can simplify as follows:

c = u^2 – z^2

And, if we rearrange this, we get:

z^2 = u^2 – c

So, we can find z by taking the square root of both sides.

square_root(z^2) = square_root(u^2 – c)
z = square_root(u^2 – c)

So, now that we have values for u and z, we can find the solution for x1 and x2.

x1 = u – z
x2 = u + z

Now, let’s try this with a specific example by setting a quadratic equation equal to zero.

x^2 – 6x + 5 = 0

To factor this equation, we want to find values for r and s where:

x^2 – 6x + 5 = 0

So, given the general form of the quadratic equation, we know that:

b = -6
c = 5

Further, since we’ve assumed that u = -b/2, we know that:
u = -(-6/2)
u = 3

We can find the value of z with this equation, which we proved above:
z^2 = u^2 – c

So:
z^2 = 3^2 – 5 = 9 – 5 = 4
z = square_root(4)
z = 2

Now we can find x1 and x2 (which will also give us r and s) as follows:

x1 = (u – z) = (3 – 2) = 1
x2 = (u + z) = (3 + 2) = 5

Therefore:

x^2 – 6x + 5 = (x – 1) (x – 5) = 0

This nice part of this is four-fold:

1. It’s easy to visualize u (i.e. the average between the two X-axis intercepts).
2. It’s also easy to visualize z (i.e. the distance between u and each x-axis intercept).
3. One can more easily see that c is where the curve intersects the y-axis.
4. One can more easily see why u = -(b/2)

Now, practically speaking we don’t really need to remember the following, but still it would be nice to know why these two statements are true:

c =  (x1) (x2)
u = -b/2

Recall that u is the average of x1 and x2.

u = (x1 + x2)/2

Note that the quadratic can be represented as follows:

x^2 + bx  +  c  =  (x + r) (x + s)

If we multiply out the right side of this equation, we get:

x^2 + bx  +  c  =  x^2 + sx + rx + rs

By combining term we get:

x^2 + bx  +  c  =  x^2 + (s + r)x + rs

From this we see that:

b = (s + r)
c = (r)(s)

When we set (x1 + r) = 0  we see that this is true when r = -x1
When we set (x2 + s) = 0  we see that this is true when s = -x2

Therefore:
b = (-x1 + -x2) = -(x1 + x2)

And:
-b = (x1 + x2)

Since u = (x1 + x2)/2 we can see that:

u = -b/2

Also, we can find c as follows:

c = (r)(s)
c = (-x1)(-x2)
c = (x1)(x2)

Here’s another example:

x^2 + 8x + 15 = 0

b = 8

u = -b/2 = -4

c = 15 = (x1)(x2) = (u – z)(u + z)
c = u^2 – z^2

z^2 = u^2    – c
z^2 = (-4)^2 – 15 = 1
z = square_root(1) = 1

x1 = (u – z)
x1 = (-4 – 1) = -5

x2 = (u + z)
x2 = (-4 + 1) = -3

(x + 5) (x + 3) = x^2 + 8x + 15

## References

Completing the Square
https://en.wikipedia.org/wiki/Completing_the_square

Mathematician Finds Easier Way to Solve Quadratic Equations