How do you simplify #sqrt(h^3)/sqrt8#?

1 Answer
Dec 14, 2017

Answer:

#+-(hsqrt(2h))/4#

Explanation:

There are two things to consider.

1) Are there any squared values we can 'take out' of the roots?

2) It is considered not good practice to have a root as or in the denominator so we need to change what is left of it into a whole number.

Given: #sqrt(h^3)/sqrt(8)#

Write as: #sqrt(h^2xxh)/(sqrt(2^2xx2))color(white)("ddd")" giving: "color(white)("ddd") (hsqrt(h))/(2sqrt(2)) #

Now we 'get rid of the root in the denominator; multiply by 1 and you do not change the inherent value. However 1 comes in many forms.

#color(green)((hsqrt(h))/(2sqrt(2))color(red)(xx1) color(white)("dddd")-> color(white)("dddd")(hsqrth)/(2sqrt2)color(red)(xxsqrt2/sqrt2) )#

#color(white)("d")color(green)(color(white)("ddddddddddd")->color(white)("dddd")(hsqrt(2h)) /4 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Foot note")#

Compare the consequence of #sqrt(h)xxsqrt(2)->sqrt(2h)color(white)("d")# to the example:

#sqrt(4)xxsqrt(9)color(white)("d")=color(white)("d")2xx3=6#

#sqrt(4xx9)color(white)("d")=color(white)("d")sqrt(36)color(white)("d")=color(white)("d")6#

#color(brown)("So " hsqrth xxsqrt2color(white)("dd") =color(white)("dd") hsqrt(hxx2)color(white)("dd") =color(white)("dd") hsqrt(2h))#

#color(white)("d")#

#color(brown)("and "2sqrt(2) xxsqrt2color(white)("d") =color(white)("d") 2sqrt(2xx2)color(white)("d") =color(white)("d") 2sqrt(4)color(white)("d") =color(white)("d") 2xx2=4#

Technically the answer should be #+-(hsqrt(2h))/4#

As the square root of a number is #+-#

Example:

#(-2)xx(-2)color(white)("d")=color(white)("d")(+4)color(white)("d")=color(white)("d")(+2)xx(+2)#