Sunday, June 29, 2014

Solving y = a*x^2 + b*x + c for x whether y is zero or not.



How to solve y = a*x^2 + b*x + c for x for when y is not necessarily zero.  We can solve for x algebraically using various methods. 

Method 1:  Completing the Square

y = a*x^2 + b*x + c

Divide both sides by a:
y/a = x^2 + b/a * x + c/a

Add  b^2/(4*a^2) to both sides. 
Note that (x + b/(2*a))^2 = x^2 + b/a * x + b^2/(4*a^2).

y/a + b^2/(4*a^2) = x^2 + b/a * x + b^2/(4*a^2) + c/a
y/a + b^2/(4*a^2) = (x + b/(2*a))^2 + c/a
y/a + b^2/(4*a^2) – c/a = (x + b/(2*a))^2

Solving for x gives:
x + b/(2*a) = ±√( y/a + b^2/(4*a^2) – c/a )
x = b/(2*a) ±√( y/a + b^2/(4*a^2) – c/a )

Method 2:  Quadratic Formula

The quadratic formula requires that the polynomial is equal to zero.  Hence:

y = a*x^2 + b*x + c
0 = a*x^2 + b*x + c – y

The constant term is c – y.  By the quadratic formula:

x = (-b ± √(b^2 – 4*a*(c-y))/(2*a)
x = (-b ± √(b^2 – 4*a*c + 4*a*y))/(2*a)

In the term with the square root, divide b^2 – 4*a*c + 4*a*y by 4*a^2:
x = (-b ± √(b^2/(4*a^2) – c/a + y/a))/(2*a)


Hope this helps,

Eddie

This blog is property of Edward Shore.  2014.

1 comment:

  1. I remember when I first derived the quadratic formula several years ago and it felt good that I could do that.

    On the last line you have ..../(2*a). Since you move that inside the square root I don't think it is needed anymore.

    ReplyDelete

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