Question

Deravation of bsquare

Answer 1

A quadratic equation (ax2 + bx + c = 0) can be solved for x by using the quadratic formula:

x = [−b ± √(b2 − 4ac)]/2a

The derivation of that formula was done by some clever manipulation of the elements in the quadratic equation. The best way to see where the formula came from is by working backwards to establish the quadratic equation. Then you can see the steps to take to derive the quadratic formula.

Questions you may have include:

What are the steps working backwards?What are the steps to get to the formula?How do you handle the square root?

This lesson will answer those questions.

Working backwards

Start with the quadratic formula and put it in the form of the quadratic equation:

x = [−b ± √(b2 − 4ac)]/2a

Multiply both sides of the equal sign by 2a

2ax = −b ± √(b2 − 4ac)

Add b to both sides

2ax + b = ± √(b2 − 4ac)

Square both sides of the equal sign

(2ax + b)2 = b2 − 4ac

4a2x2 + 4abx + b2 = b2 − 4ac

Subtract b2 from both sides

4a2x2 + 4abx = − 4ac

Add − 4ac to both sides

4a2x2 + 4abx + 4ac = 0

Divide by 4a

ax2 + bx + c = 0

Start from equation

Knowing the steps, you can start from the quadratic equation to get the formula:

ax2 + bx + c = 0

Multiply both sides of the equal sign by 4a

4a2x2 + 4abx + 4ac = 0

4a2x2 + 4abx = − 4ac

4a2x2 + 4abx + b2 = b2 − 4ac

Factor 4a2x2 + 4abx + b2 to get (2ax + b)2

(2ax + b)2 = b2 − 4ac

Taking the square root

When you get to (2ax + b)2 = b2 − 4ac, you want to take the square root of each side of the equal sign.

Note that both (2ax + b)*(2ax + b) and [−(2ax + b)]*[−(2ax + b)] equal (2ax + b)2.

That means that the square root can be either plus (+) or minus (−). Thus:

x = [−b ± √(b2 − 4ac)]/2a

The derivation is complete.

Summary

You can see the steps to derive the quadratic formula 
x = [−b ± √(b2 − 4ac)]/2a by first going backwards to get the quadratic equation (ax2 + bx + c = 0). Then you can reverse the steps to go from the equation to the formula. When you take the square root, you need to realize that plus or minus factors can be used.