saifoo.khan
  • saifoo.khan
Is this true? Why?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Is what true?
saifoo.khan
  • saifoo.khan
\[\Large \sqrt{a^2} = (\sqrt{a})^2\]
anonymous
  • anonymous
true :)

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More answers

saifoo.khan
  • saifoo.khan
And why?
anonymous
  • anonymous
for example: \[\sqrt{9} = 3 = (\sqrt{9})^{2}\]
anonymous
  • anonymous
Both sides are equal, that's why it's true.
saifoo.khan
  • saifoo.khan
Are you guys sure? xD
anonymous
  • anonymous
enothing is %100 ! :D
austinL
  • austinL
\(\sqrt{4^2} = \sqrt{16}=4\) \((\sqrt{4})^2=4\)
saifoo.khan
  • saifoo.khan
What if we take -4? @austinL
anonymous
  • anonymous
\[\pm \sqrt{x}\]
anonymous
  • anonymous
notm\[\not +\sqrt{x} \]
inkyvoyd
  • inkyvoyd
it's not true - for -4 you get 4 and 4i
austinL
  • austinL
\(\sqrt{(-4)^2}=\sqrt{16}=4\) \((\sqrt{(-4)})^2=-4\) It doesn't necessarily hold true in all instances.
inkyvoyd
  • inkyvoyd
no, *-4
anonymous
  • anonymous
but @austinL I think the second one is false!
inkyvoyd
  • inkyvoyd
how about taking x=4i?
austinL
  • austinL
But, the square would remove the square root first before you evaluate the root. \(((-4)^{\frac{1}{2}})^2=-4\)
anonymous
  • anonymous
yes,so it is false :D
austinL
  • austinL
Exactly, so if one instance breaks the statement, then you must hold it false. How could we make this true? ;)
anonymous
  • anonymous
I will say this to my teacher,Thank you :) :D
inkyvoyd
  • inkyvoyd
guys guys take x=4i :3

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