A not so New way to solve Quadratic Equations

Question

Well I came across a new way to solve quadratic equations. It works on equations which are hard to factor. I read about it on New York Times. in this parabola:
 
y = x2 – 4x – 5
 
The two solutions when y = 0 are the symmetrical points r and s, where the parabola crosses the x-axis.
 
The midpoint, or average, of r and s is the axis of symmetry of the parabola. We want r + s = –b, which happens when the average of r and s is –b ÷ 2. In this example: 4 ÷ 2 = 2.
The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example:
 
r = 2 – u and s = 2 + u
 
To find u, we want the product of r and s to be equal to c, which in this example is –5. Rewriting r and s in terms of u:
 
r × s = –5
(2 – u) × (2 + u) = –5
 
Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
 
The two solutions to this quadratic equation are 2 – u and 2 + u, or –1 and 5. In other words, this parabola intersects the x-axis when x = –1 and x = 5.

bhaskar123 · Answer

On Second read i notice u^2 =9 so u=3 or -3 as GMAT sees it, but know that u here is the distance so u got to have a positive root.

A not so New way to solve Quadratic Equations

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