How do you simplify #sqrt(63c^3d^4f^5)#?

2 Answers
Mar 21, 2017

Answer:

#3cd^2f^2*sqrt(7cf)#

Explanation:

#sqrt(63c^3d^4f^5)#

#:.=(63c^3d^4f^5)^(1/2)#

#:.=(7*3*3*c^3d^4f^5)^(1/2)#

#:.=(7*3^2c^3d^4f^5)^(1/2)#

#:.=(7^(1/2)*3^(2 xx 1/2)c^(3 xx 1/2)d^(4 xx 1/2)f^(5 xx 1/2))#

#:.=(7^(1/2)*3^1*c^(3/2)*d^(4/2)*f^(5/2))#

#:.=sqrt7*3*sqrt(c^3)*sqrt(d^4)*sqrt(f^5)#

#:.=3d^2*sqrt(7c^3f^5#

#:.=3d^2sqrt(7*c*c*c*f*f*f*f*f)#

#sqrta*sqrta=a#

#:.=3cd^2f^2sqrt(7cf)#

Mar 21, 2017

Answer:

#3cd^2f^2sqrt(7cf)#

Explanation:

Given#:" "sqrt(63c^3d^4f^5)#

As soon as you see a square root you know that you are looking for squared values within that root. You can 'take outside' the root all those squared values.

Write as:#" "sqrt( 3^2xx7xxc^2xxcxx(d^2)^2xx(f^2)^2xxf)#

Taking the squared values 'outside' the root giving:

#3cd^2f^2sqrt(7cf)#