square roots: find square root of 9x^2 - 6x + 1

[matt123]

How do you find the square root here?

9x^2 - 6x + 1

and or here?

9x^2/y^2 + 18 + 9y^2/x^2

thanks.

 

[galactus]

Re: square roots

Find the square root?. Do you want to factor the quadratic?.

 

[matt123]

Re: square roots

I think so, is that generally what you do for this kind of thing?

 

[Denis]

Re: square roots

Matt, post an ORIGINAL problem in FULL; else not much we can do...

 

[matt123]

Re: square roots

Matt, post an ORIGINAL problem in FULL; else not much we can do...

Hmm, these are original problems, it says find the square root and thats it.
Do you want another problem?
Here is a example x^2 + 6x + 9 = x+3
I don't get this either though.

 

[galactus]

Re: square roots

I don't know what they mean by find the square roots either. But, we can factor. I will show this one and you try the other.

$\displaystyle x^{2}+6x+9=x+3$

$\displaystyle x^{2}+5x+6=0$

What two numbers when multiplied equal 6 and when added equal 5?. Hmmmm....how about 2 and 3.

$\displaystyle (x+2)(x+3)$

So, the solutions are x=-2 and x=-3

 

[skeeter]

Re: square roots

How do you find the square root here?

note that you need to know how to factor an expression that is a perfect square ...

9x^2 - 6x + 1

$\displaystyle \sqrt{9x^2 - 6x + 1} = \sqrt{(3x+1)^2} = |3x+1|$

and or here?
9x^2/y^2 + 18 + 9y^2/x^2

$\displaystyle \sqrt{\frac{9x^2}{y^2} + 18 + \frac{9y^2}{x^2}} =$
$\displaystyle \sqrt{9 \left(\frac{x^2}{y^2} + 2 + \frac{y^2}{x^2}\right)} =$
$\displaystyle \sqrt{9 \left( \frac{x}{y} + \frac{y}{x} \right)^2} = 3 \left \lvert \frac{x}{y} + \frac{y}{x} \right \rvert$

 

[matt123]

Thanks for the help.