How do you simplify 4 * the square root of 7 + 8 * the square root of 63?

3 Answers
Apr 1, 2018

Answer:

#28sqrt7#

Explanation:

#"using the "color(blue)"law of radicals"#

#•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)#

#rArrsqrt63=sqrt(9xx7)=sqrt9xxsqrt7=3sqrt7#

#rArr4sqrt7+(8xx3sqrt7)#

#=4sqrt7+24sqrt7#

#=28sqrt7#

Apr 1, 2018

Answer:

#28sqrt7#

Explanation:

Here,

#4sqrt7+8sqrt63=4sqrt7+8sqrt(9xx7#

#=4sqrt7+8*3sqrt7#

#=4sqrt7+24sqrt7#

#=28sqrt7#

Apr 1, 2018

Answer:

#28sqrt(7)#

Explanation:

Assumption: You mean #color(white)("d")4sqrt(7)+8sqrt(63)#

7 is a prime number so lets leave that for the moment.

Consider #sqrt(63)#

Useful trick follows.
note that for 63 #->6+3=9# and as 9 is divisible by 3 then so is 63

#sqrt(63) =sqrt(3xx21) = sqrt(3xx3xx7) larr" The 7 is useful!"#
#color(white)("dd...")=3sqrt(7)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all back together we have:

#4sqrt(7)+(8xx3sqrt(7))#

#4sqrt(7)+24sqrt(7)#

Factor out the #sqrt(7)#

#sqrt(7)(4+24)#

#28sqrt(7)#