anonymous
  • anonymous
how would you solve (sqrt(x+2)) + (sqrt(3-x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sqrt{x+2}+\sqrt{3-x}=0 ?\]
anonymous
  • anonymous
\[\sqrt{a} ~~~\,a \ge0\]
anonymous
  • anonymous
\[x+2\ge0 \,and \,~3-x \ge0\]

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anonymous
  • anonymous
f(x) = \[\sqrt{x+2}\] ; g(x) = \[\sqrt{3-x}\] find (f+g)(x) how would I do that?
anonymous
  • anonymous
\[x \ge2~\, and~\,x \le3\]
anonymous
  • anonymous
ohh ok
anonymous
  • anonymous
(f+g)(x)=f(x)+g(x) \[(f+g)(x)=\sqrt{x+2}+\sqrt{3-x}\] this is the answer you cant do anything
anonymous
  • anonymous
what if it was (f * g) ? could you do something then?
anonymous
  • anonymous
since inside of the roots are different you cannot make addition
anonymous
  • anonymous
is it multiply sign?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
if it was (f*g)(x), you can multiply the expression inside the radical signs since they have common indices.
anonymous
  • anonymous
(f*g)(x)=f(x)*g(x) \[(f*g)(x)=(\sqrt{x+2})(\sqrt{3-x})\] \[\sqrt{(x+2)(3-x)}\]\[\sqrt{-x^2+x-6}\]
anonymous
  • anonymous
it should be +6 not -6
anonymous
  • anonymous
o okay that makes sense thank you
anonymous
  • anonymous
yw

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