## 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

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 )

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.

### Scientific Calculator Basics Document Now Available

Scientific Calculator Basics Document Now Available If you are new to scientific calculators, ever wondered what the [sin], [log],...