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?

1 Answer
Oct 24, 2015

Answer:

#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.