Question

If the quadratic equation can be used to determine when a function equals zero, is there a modified quadratic equation that can determine when a function equals another constant?

Answer 1

`x = (-b+-sqrt(b^2 +4a(q-c)))/(2a)`

Explanation:

Any time you want to find for what value of `x` a function is equal to a constant, you just set the function equal to that constant and solve for `x`. Quadratic functions are no different. The quadratic equation is derived by setting a quadratic function equal to zero and solving for `x`.

We can follow the original derivation but set the function to the desired constant instead of `0`. Lets call that constant `q`. The original derivation of the quadratic formula can be found here.

We will start by setting the general quadratic function equal to `q`.

`ax^2 +bx+c = q`

Move `c` to the right hand side.

`ax^2+bx=q-c`

Now we divide by `a`.

`x^2 + b/ax = (q-c)/a`

At this point we need to complete the square. Here is an explanation of what that means. We add `b^2/(4a^2)` to each side in order to complete the sqare.

`x^2 +b/ax +b^2/(4a^2) = (q-c)/a +b^2/(4a^2)`

Now the left hand side is in a form that can be simplified into a "neat" square.

`(x + b/(2a))^2=(q-c)/a+b^2/(4a^2)`

At this point we want a common denominator on the right hand side in order to combine both fractions together. Multiply the first fraction by `(4a)/(4a)`.

`(x + b/(2a))^2=(4a(q-c))/(4a^2)+b^2/(4a^2)=(b^2 +4a(q-c))/(4a^2)`

I rearranged the right hand side a little bit in order to make it look more like the original quadratic equation. We can now take the square root of both sides.

`x + b/(2a)=+-sqrt((b^2 +4a(q-c))/(4a^2))=+-sqrt(b^2 +4a(q-c))/(2a)`

Lastly, move the constant term from the left to the right by subtracting from both sides.

`x = (-b+-sqrt(b^2 +4a(q-c)))/(2a)`

There you have it, the only difference from the original quadratic formula is that the `-c` becomes `q-c`. If you set `q` equal to `0`, you get the original quadratic formula.