Completing the Square

[nycmathguy]

We need to complete the square when an equation cannot be factored. Yes?

 

[MathLover1]

When asked to solve a quadratic equation that you just can't seem to factor (or that just doesn't factor), you have to employ other ways of solving the equation, such as by using the quadratic formula. The quadratic formula is the formula used to solve for the variable in a quadratic equation in standard form.

when to use "completing the square"
It depends on what information you are trying to get and how simple the quadratic problem you are facing is.
If you are trying to find the vertex of a parabola described by a quadratic equation, then completing the square is the most natural way to do it.
If you are trying to find the roots of a quadratic equation, then completing the square will 'always work', in the sense that it does not require the factors to be rational and in the sense that it will give you the complex roots if the quadratic's roots are not real.
On the other hand, there may be obvious or easy to find factorings that are a little quicker.

In general you can complete the square as follows:

I usually first check Δ=b^2−4ac to see if I'm facing a quadratic that will factor nicely or I have to use heavier methods.

 

[nycmathguy]

When asked to solve a quadratic equation that you just can't seem to factor (or that just doesn't factor), you have to employ other ways of solving the equation, such as by using the quadratic formula. The quadratic formula is the formula used to solve for the variable in a quadratic equation in standard form.

when to use "completing the square"
It depends on what information you are trying to get and how simple the quadratic problem you are facing is.
If you are trying to find the vertex of a parabola described by a quadratic equation, then completing the square is the most natural way to do it.
If you are trying to find the roots of a quadratic equation, then completing the square will 'always work', in the sense that it does not require the factors to be rational and in the sense that it will give you the complex roots if the quadratic's roots are not real.
On the other hand, there may be obvious or easy to find factorings that are a little quicker.

In general you can complete the square as follows:

I usually first check Δ=b^2−4ac to see if I'm facing a quadratic that will factor nicely or I have to use heavier methods.

What do you mean by triangle = b^2 - 4ac?
Note: There are some changes that had to be done concerning our precalculus trek. Look for my PM (private message).

 

[Country Boy]

You just said "must use the quadratic formula" and now are saying "must complete the square"! No, you can use either one.

(The quadratic formula is derived by completing the square. Given ax^2+ bx+ c= 0, divide through by a [in order that this be quadratic a must not be 0]: x^2+ (b/a)x+ c/a= 0.

Subtract c/a from both sides: x^2+ (b/a)x= -c/a.

Complete the square by adding b^2/4a^2 to both sides: x^2+ (b/a)x+ b^2/4a^2= b^2/4a^2- c/a
(x+ b/2a)^2= (b^2- 4ac)/4a^2

Take the square root of both sides:
x+ b/2a= +/- sqrt(b^2- 4ac)/2a

Subtract b/2a from both sides:
x= (-b+/- sqrt(b^-2- 4ac)/2a.

 

[MathLover1]

What do you mean by triangle = b^2 - 4ac?
Note: There are some changes that had to be done concerning our precalculus trek. Look for my PM (private message).

Uppercase delta (Δ) in algebra represents the discriminant of a polynomial equation. This polynomial equation is almost always the quadratic equation.
Consider the quadratic ax^2+bx=c, the discriminant of this equation would equal b^2-4ac, and it would certainly look like this: Δ= b^2-4ac.

 

[nycmathguy]

You just said "must use the quadratic formula" and now are saying "must complete the square"! No, you can use either one.

(The quadratic formula is derived by completing the square. Given ax^2+ bx+ c= 0, divide through by a [in order that this be quadratic a must not be 0]: x^2+ (b/a)x+ c/a= 0.

Subtract c/a from both sides: x^2+ (b/a)x= -c/a.

Complete the square by adding b^2/4a^2 to both sides: x^2+ (b/a)x+ b^2/4a^2= b^2/4a^2- c/a
(x+ b/2a)^2= (b^2- 4ac)/4a^2

Take the square root of both sides:
x+ b/2a= +/- sqrt(b^2- 4ac)/2a

Subtract b/2a from both sides:
x= (-b+/- sqrt(b^-2- 4ac)/2a.

Very cool.

 

[nycmathguy]

Uppercase delta (Δ) in algebra represents the discriminant of a polynomial equation. This polynomial equation is almost always the quadratic equation.
Consider the quadratic ax^2+bx=c, the discriminant of this equation would equal b^2-4ac, and it would certainly look like this: Δ= b^2-4ac.

I confused your delta with delta y ÷ delta x. I will reply to your PM later on my break at the job. Learning the WHY things occur in math is cool but did you forget about my overnight schedule? I cannot dive into math the same way a retired person does. More about this through PM. Check out my two unanswered questions in the Geometry and Trigonometry forum here. Thanks for your daily help and guidance.