anonymous
  • anonymous
√162x^3(y^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[√162x^3y^2\]
anonymous
  • anonymous
idk sorry
anonymous
  • anonymous
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mathstudent55
  • mathstudent55
Is this the problem? \(\sqrt{162x^3y^2}\)
anonymous
  • anonymous
yes
anonymous
  • anonymous
the answer is \[9xy \sqrt{2x}\]
mathstudent55
  • mathstudent55
\(\sqrt{162x^3y^2}\) \(= \sqrt{81 \times 2 \times x^2 \times x \times y^2}\) \(= \sqrt{81}\sqrt{x^2y^2}\sqrt{2x}\) \(= 9\sqrt{x^2y^2}\sqrt{2x}\) The root of 81 is 9. That is simple. The roots of 2 and x are irrational and stay inside the root. That is also simple. The question is what to do with the roots of x^2 and y^2.
mathstudent55
  • mathstudent55
If the problem told you to assume the variables are non-negative, then \(\sqrt{x^2} = x\) and \(\sqrt{y^2} = y\) If you were not told these variables are non-negative, then the roots are: \(\sqrt{x^2} = |x|\) and \(\sqrt{y^2} = |y|\)
mathstudent55
  • mathstudent55
If you were told the variables are non-negative, then you have: \(= 9xy\sqrt{2x}\) If you were not told the variables are non-negative, then the answer is: \(= 9|xy|\sqrt{2x}\)

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