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

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matt123

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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.
 
Re: square roots

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

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

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

Denis said:
Matt, post an ORIGINAL problem in FULL; else not much we can do...
Click to expand...
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.
 
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
 
Re: square roots

matt123 said:
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

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