THE QUADRATIC FORMULA It works for everything! …as long as it’s a quadratic equation.

Sick of all these different methods?  Reverse FOIL  With a coefficient on x 2  Without one  Completing the Square  Answers with square roots left in them  Answers with imaginary numbers

What are we going to do about it?  Let’s look at the most general form of quadratic formula and solve that for x.  That way we can just plug in every time for “x=“ (since that’s what we’re looking for).  What is the most general form of the quadratic equation?  ax 2 + bx + c = 0

Deriving the quadratic formula  ax 2 + bx + c = 0  Well, normally to solve for x we would factor…but if we don’t know what the numbers are, we can’t factor.  What do you do if you can’t factor?  Complete the square!

Deriving quadform (continued)  ax 2 + bx + c = 0  First step in completing the square – move the constant over.  ax 2 + bx = -c  Now before we do the next step, let’s divide everything by a – so we don’t have to mess with the really complicated formula for what we add to both sides.

Deriving quadform (continued)  Okay, the thing we add to both sides is what?  Half the x-term, squared.  The whole point of this is to factor it to (x + #) 2

Deriving quadform (still)  Okay, now it gets interesting.  Before we take the square roots of both sides, I’m going to multiply everything by 4a 2. Do I have to?  No, but it’ll simplify in fewer steps if I do.  So:

Deriving quadform (continued)  Multiply through where possible. Now take the square root of both sides.  Multiply through where possible.

Deriving quadform (the end)  Now we rearrange to solve for x  Subtract b from both sides  Divide by 2a  And we’re done!

So how do we use this?  Just plug in the coefficients – a, b, and c – into the equation.  EX:  x 2 + 8x = 11  x 2 + 8x – 11= 0  a = 1  b = 8  c = -11  “Plug and Chug,” as they say.

Example Solution  x = (-8 +\- √(8 2 – 4(1)(-11)))/(2(1))  x = (-8 +\- √( ))/2  x = (-8 +\- √(108))/2  Ok, have to use square root knowledge here  108 = 12*9 = 4*3*9 = 4*9*3, so  √(108) = √(4*9*3) = 2*3* √(3) = 6 √(3)  x = (-8 +\- 6 √(3))/2  x = √(3), -4 – 3√(3)