quadratic formula

Question

Discuss how each of the methods (completing the square, and factoring) can actually be easier and more efficient for solving certain types of quadratic equations?

 

Explain why it is important to be exposed to the first ways before you can fully understand the second way(quadratic formula)?

Answer 1

Good explanation.....Pick the quad. formula and stick with it. It works all the time.

Get GOOD at it.

Answer 2

Deriving the quadratic formula quickly requires you to complete the square or in the first place.

 

As for efficiency, here are the reasons for each alternative method.

 

Factoring:

It is easy to factor a quadratic or to prove it isn't factorable. In ax²+bx+c=0, all you have to do is find two integers m and n such that mn=ac and m+n=b. Then you can write the quadratic as ax²+mx+nx+c=0. Since a:m=n:b, you can quickly factor by grouping. The advantage here is that you will immediately recognize that the two roots are rational, which is not always obvious from the quadratic formula. For example, if b²-4ac=4489, it is not obvious that 4489=67² since the number is quite large.

 

Completing the square:

This helps with keeping the numbers small since before doing this you would first divide by the leading coefficient. Since you are taking square roots of smaller numbers after having to have completed the square, you have less simplifying to do.

 

As to my first statement, here is the derivation of the quadratic formula.

 

ax² + bx + c = 0   The original quadratic.

 

x² + (b/a)x + c/a = 0   Divide by a.

 

x² + (b/a)x = -c/a    Move constant term.

 

x² + (b/a)x +b²/(4a²) = b²/(4a²) - c/a    Complete the square.

 

[x+b/(2a)]² = (b²-4ac)/(2a)²       Factor left side/simplify the right side.

 

x+b/(2a) = ±√(b²-4ac)/(2a)       Take square roots.

 

x = [-b±√(b2-4ac)]/(2a)           Move constant term.