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.
