Square Root Of 9x 2
matt123
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- #1
How do y'all discover the foursquare root here?
9x^2 - 6x + one
and or here?
9x^2/y^ii + 18 + 9y^ii/x^ii
cheers.
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- #two
Re: square roots
Find the square root?. Do yous want to factor the quadratic?.
matt123
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- #iii
Re: square roots
I think so, is that generally what y'all practice for this kind of affair?
Denis
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Re: foursquare roots
Matt, mail service an ORIGINAL problem in Full; else not much we can practise...
matt123
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Re: square roots
Denis said:
Hmm, these are original bug, it says find the square root and thats it.
Practise you desire another problem?
Here is a example x^2 + 6x + 9 = ten+3
I don't go this either though.
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Re: foursquare roots
I don't know what they mean by observe the square roots either. But, nosotros can cistron. I will prove this one and y'all try the other.
\(\displaystyle x^{2}+6x+nine=ten+three\)
\(\displaystyle ten^{2}+5x+six=0\)
What 2 numbers when multiplied equal half-dozen and when added equal v?. Hmmmm....how near 2 and 3.
\(\displaystyle (x+2)(x+3)\)
And then, the solutions are x=-two and x=-3
skeeter
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- #vii
Re: foursquare 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
\(\displaystyle \sqrt{9x^2 - 6x + i} = \sqrt{(3x+1)^2} = |3x+ane|\)
and or here?
9x^two/y^2 + 18 + 9y^2/x^2\(\displaystyle \sqrt{\frac{9x^two}{y^2} + 18 + \frac{9y^2}{x^2}} =\)
\(\displaystyle \sqrt{ix \left(\frac{x^ii}{y^2} + two + \frac{y^2}{x^ii}\correct)} =\)
\(\displaystyle \sqrt{9 \left( \frac{x}{y} + \frac{y}{x} \right)^2} = 3 \left \lvert \frac{x}{y} + \frac{y}{x} \right \rvert\)
matt123
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Square Root Of 9x 2,
Source: https://www.freemathhelp.com/forum/threads/square-roots-find-square-root-of-9x-2-6x-1.56670/
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