How do you simplify #7sqrt3*2sqrt6#?

3 Answers
May 7, 2017

Answer:

#42sqrt2#

Explanation:

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(sqrtaxxsqrtbhArrsqrt(axxb))color(white)(2/2)|)))#

#rArr7sqrt3xx2sqrt6#

#=7xx2xxsqrt3xxsqrt6#

#=14xxsqrt(3xx6)#

#=14xxsqrt18#

#["now " sqrt18=sqrt(9xx2)=sqrt9xxsqrt2=3sqrt2]#

#=14xx3sqrt2#

#=42sqrt2#

May 7, 2017

Answer:

See a solution process below:

Explanation:

First, we can rewrite this expression as:

#(7 * 2)(sqrt(3) * sqrt(6)) -> 14(sqrt(3) * sqrt(6))#

We can now use this rule for multiplying radicals to continue the simplification:

#sqrt(a) * sqrt(b) = sqrt(a * b)#

#14(sqrt(3) * sqrt(6) => 14 sqrt(3 * 6) => 14sqrt(18)#

We can now use the same rule we used above but in reverse to rewrite this expression as:

#14sqrt(18) => 4sqrt(9 * 2) => 4(sqrt(9) * sqrt(2)) => 14(+-3 * sqrt(2)) =>#

#+-42sqrt(2)#

May 7, 2017

#sqrt(a.b) = sqrta.sqrtb#

so, #sqrt6 = sqrt3.sqrt2#

Then, #7sqrt3.(2sqrt6) = 7sqrt3.(2sqrt3).sqrt2#

i.e. #14(3).sqrt2#

so, #42sqrt2#

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