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Mathematics 8 Online
OpenStudy (anonymous):

solve x^2 +12x + 5 = 0

OpenStudy (anonymous):

x = {12.403124237, -0.403124237}

OpenStudy (jhannybean):

You want to use the method of completing the square, are you familiar with that @thefourdcarrs ?

OpenStudy (anonymous):

Yes except that there are no multiples of 5 that add up to 12 so would I have to use the quadratic equation formula?

OpenStudy (jhannybean):

Yes, the quadratic formula, or it's accomplice, the completing the square method.

OpenStudy (jhannybean):

Using this method, you are going to "form a quadratic function" inside the parenthesis.

OpenStudy (anonymous):

Not sure how to do it by using completing the square method

OpenStudy (jhannybean):

You shall learn it today! :)

OpenStudy (anonymous):

I appreciate your help

OpenStudy (jhannybean):

So first, group your x terms together like so: \((x^2+12x) +5=0\)

OpenStudy (jhannybean):

comparing the stuff inside the parenthesis to the quadratic formula, \(ax^2 +bx+c\), we see that we have part of it completed, the \(ax^2 +bx\) part. Do you see that?

OpenStudy (anonymous):

Yes

OpenStudy (jhannybean):

Alright :) Now we want to compete the quadratic, so we need to find a \(c\) value that will satisfy the equation. We can find that by taking our \(b\) value, dividing it by 2, and then squaring the whole fraction as so: \(c= \left(\frac{b}{2}\right)^2\)

OpenStudy (jhannybean):

Could you find the new c value for me?

OpenStudy (anonymous):

not sure where to go from that part though factoring out the x (x+12) +5 = 0?

OpenStudy (jhannybean):

Not quite. Remember, we have part of our quadratic completed. therefore our \(b\) is 12 in this case.\[c=\left(\frac{12}{2}\right)^2 = 6^2=36\]

OpenStudy (jhannybean):

Then we input this back in to our equation. \[(x^2+12x+36) =-5+36\]What we add to one side of our equation, we must subtract to keep our reformatted equation consistent with our original equation. I isolated our new found quadratic on the left side by subtracting -5 to both sides, and then added in the \(c\) value of our quadratic to both sides to stay consistent.

OpenStudy (fibonaccichick666):

If it's not dimple and doesn't yield an imaginary answer, I prefer just using the quadrac formula. Factoring(ie completing the square, which I hate) is my personal last resort.

OpenStudy (jhannybean):

Oh, I love it!! Haha.

OpenStudy (jhannybean):

Quadratics are my last resort, haha.

OpenStudy (fibonaccichick666):

simple* lol

OpenStudy (fibonaccichick666):

It's too much effort to do completing the square haha

OpenStudy (jhannybean):

I've been rigorously practicing this method since coming to openstudy, and now I find it wonderfully simple :P

OpenStudy (fibonaccichick666):

lol I just don't like it. I prefer a neat result.

OpenStudy (fibonaccichick666):

that isn't neat to me

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