Can anyone help me, for solve this?Please,thank u!

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3 Answers
Feb 23, 2018

See explanation...

Explanation:

Hello!

I noticed this is your first post here on Socratic, so welcome!!

Just looking at this problem, we know right off the bat that we need to somehow get rid of the "squares." We also know that you can't square an #8# Notice that one #x^2# is negative, which normally means we should move it to the other side. Let me explain:

#x^2=8-x^2#

Move the #x^2# to the other side by adding it to both sides

#x^2+x^2=8 cancel(-x^2)cancel(+x^2)#

#2x^2=8#

Divide both sides by #2#

#(cancel2x^2)/cancel2=8/2#

#x^2=4#

Finally, take the square root to get rid of the square

#cancelsqrt(x^cancel2)=sqrt4#

#x=+-2#

Hope this helped!
~Chandler Dowd

Feb 23, 2018

X=+2 or -2

Explanation:

Given,#y^2=X^2#--------#color(red)(1)#
And,#y^2=8-X^2#
Or,#X^2=8-X^2# #color(blue){ (From 1)}#
Or,#X^2=8//2 #
Thus,#X=sqrt(4)#
#X=+-2#

Feb 23, 2018

Ok so first of all I'm french so sorry if I don't use the proper terms I'm still learning.

#x^2 = 8 - x^2#

First you have to put all x's on the same side. Don't forget to invert the signs !

#x^2 + x^2 = 8#. It's the same as #2x^2 = 8#.

In order to find the value of x, you have to get rid of all this fancy crap around it by dividing stuff. It's not very clear when I put it like this, so here goes :

#2x^2-:2x=x#. In order to get the value of your now found x, you just have to do the same on the other side !

You get #8-:2x# which I believe you just have to write as a decimal number since you can't really reduce it anymore.

Anyways that's, I hope, the correct result. If somebody could double check though, that would be great.

Hope this helps !