# 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?

Oct 24, 2015

$x = \frac{- b \pm \sqrt{{b}^{2} + 4 a \left(q - c\right)}}{2 a}$

#### 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 $q$.

$a {x}^{2} + b x + c = q$

Move $c$ to the right hand side.

$a {x}^{2} + b x = q - c$

Now we divide by $a$.

${x}^{2} + \frac{b}{a} x = \frac{q - c}{a}$

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

${x}^{2} + \frac{b}{a} x + {b}^{2} / \left(4 {a}^{2}\right) = \frac{q - c}{a} + {b}^{2} / \left(4 {a}^{2}\right)$

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

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{q - c}{a} + {b}^{2} / \left(4 {a}^{2}\right)$

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 $\frac{4 a}{4 a}$.

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{4 a \left(q - c\right)}{4 {a}^{2}} + {b}^{2} / \left(4 {a}^{2}\right) = \frac{{b}^{2} + 4 a \left(q - c\right)}{4 {a}^{2}}$

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.

$x + \frac{b}{2 a} = \pm \sqrt{\frac{{b}^{2} + 4 a \left(q - c\right)}{4 {a}^{2}}} = \pm \frac{\sqrt{{b}^{2} + 4 a \left(q - c\right)}}{2 a}$

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

$x = \frac{- b \pm \sqrt{{b}^{2} + 4 a \left(q - c\right)}}{2 a}$

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.