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If x>0, what is the quotient of 20āˆš z^6/āˆš 16z^7 in simplest radical form?

Sagot :

I suppose you meant z > 0, or meant to use x's in place of z'sā€¦

If z > 0, then

[tex]\dfrac{20\sqrt{z^6}}{\sqrt{16z^7}} = \dfrac{20\sqrt{\left(z^3\right)^2}}{\sqrt{4^2\left(z^3\right)^2z}} = \dfrac{20\left|z^3\right|}{4\left|z^3\right|\sqrt{z}} = \boxed{\dfrac{5}{\sqrt{z}}}[/tex]

because

[tex]\sqrt{z^2} = |z|[/tex]

and if z is positive, we have |z| = z, and we can eliminate common factors of z in the fraction,

[tex]\dfrac{\left|z^3\right|}{\left|z^3\right|} = \dfrac{z^3}{z^3} = 1[/tex]