Solve equation in quadratic form

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I'm getting pretty rusty at algebra and cant' remember how to do this.

My problem says \(\displaystyle Solve -6+1/x^2=5/x\) in quadratic form.

I simplified the problem to get -6x^2 + 1 = 5x. (I hope I did this step right)

Then I used the quadratic formula to get

\(\displaystyle x = (-5\)± \(\displaystyle sqrt1) / -2\)



So basically, I just need someone to make sure I did this correctly.

Thank you in advance!
 
No, that's not correct. Your quadratic is OK, though.

Factor instead.

\(\displaystyle \L\\{-}6x^{2}-5x+1=0\)

What two numbers when added equal -5 and when multiplied equal -6

-6 and 1.

\(\displaystyle \L\\-6x^{2}-5x+1=-6x^{2}-6x+x+1=(-6x^{2}-6x)+(x+1)\)

Factor:

\(\displaystyle \L\\(-6x^{2}-6x)+(x+1)=-6x(x+1)+(x+1)=(-6x+1)(x+1)\)

See the solutions now?.
 
I'm not understanding how you get this:

\(\displaystyle \L\\-6x^{2}-5x+1=-6x^{2}-6x+x+1=(-6x^{2}-6x)+(x+1)\)

Where does the 6x^2 - 6x + x + 1 come from?
 
-6x+x equals -5x. doesn't it?. You've not learned to factor a quadratic?.
 
catalinamemday said:
No, I don't remember ever [factoring quadratics].
Click to expand...
This is a necessary skill, and the topic will arise in other contexts in later classes, wherein it will be assumed that you know how to factor. It would probably be a good idea to study some online lessons now, so you'll have the skill for later.

Eliz.
 
Well, just because I am one heck of a nice guy, I will give you a quick run through. If you have taken algebra, you should've learned this.

I would rather factor, if possible, then use the quadratic formula.

If faced with a quadratic such as:

\(\displaystyle \L\\x^{2}+x-2=0\)

See the coefficient of x is 1 and the constant term is -2.

We find two numbers which when added equal 1 and when multiplied equal -2

That would be 2 and -1. Because (2)(-1)=-2 and 2+(-1)=1

Sub them in for x because 2x-x=x

\(\displaystyle \L\\x^{2}+2x-x-2\)

Group:

\(\displaystyle \L\\(x^{2}+2x)-(x+2)\)

Factor out common terms in each:

\(\displaystyle \L\\x(x+2)-(x+2)\)

\(\displaystyle \L\\(x-1)(x+2)\)

Therefore, the solutions are 1 and -2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Now, when the coefficient of x^2 is something other than 1. In your case -6. You must take that into account, as I done previously.

\(\displaystyle \L\\-6x^{2}-5x+1\)

Solve:

\(\displaystyle \L\\a+b=-5\)
\(\displaystyle \L\\ab=-6\)

This gives a=1 or -6 and b=1 or -6

Therefore, you have:

\(\displaystyle \L\\-6x^{2}-6x+x+1\)

\(\displaystyle \L\\(-6x^{2}-6x)+(x+1)\)

\(\displaystyle \L\\-6x(x+1)+(x+1)\)

\(\displaystyle \L\\(-6x+1)(x+1)\)

The solutions are 1/6 and -1

As I showed before.

If you need anymore, you can find plenty online. Try http://www.Purplemath.com
 
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