How do you multiply #sqrt(2)(sqrt 8 + sqrt 4)#?

2 Answers
Jun 29, 2015

Answer:

#sqrt(2)(sqrt(8)+sqrt(4)) = 4+2sqrt(2)#

Explanation:

Method 1: Multiply first then simplify
The distributive property (of multiplication over addition) tells us that
#color(white)("XXXX")##a*(b+c) = ab+ac#

Further as a property of exponents (and therefore of square roots)
#color(white)("XXXX")##a^m*b^m = (a*b)^m#
#color(white)("XXXX")##sqrt(a)*sqrt(b) = sqrt(a*b)# since #(sqrt(k) = k^(1/2))#

So
#sqrt(2)(sqrt(8)+sqrt(4))#
#color(white)("XXXX")##=sqrt(2)*sqrt(8) + sqrt(2)*sqrt(4)#

#color(white)("XXXX")##=sqrt(2*8) + sqrt(2*4)#

#color(white)("XXXX")##=sqrt(16) + sqrt(8)#

#color(white)("XXXX")##=4+ 2sqrt(2)#

Method 2: Simplify roots then multiply
#sqrt(2)(sqrt(8)+sqrt(4))#
#color(white)("XXXX")##=sqrt(2)(2sqrt(2)+2)#

#color(white)("XXXX")##=2sqrt(2)*sqrt(2) + 2sqrt(2)#

#color(white)("XXXX")##=2*2 + 2sqrt(2)#

#color(white)("XXXX")##=4+2sqrt(2)#

Jun 29, 2015

Answer:

#sqrt(2)(sqrt 8 + sqrt 4)=2(2+sqrt(2))#

Explanation:

  1. #sqrt(2)(sqrt 8 + sqrt 4)=sqrt(2)*sqrt(8)+sqrt(2)*sqrt(4)# for the distributive property.
  2. #sqrt(2)*sqrt(8)+sqrt(2)*sqrt(4)=sqrt(2*8)+sqrt(2*4)=sqrt(2^4)+sqrt(2^3)# because we can write the product of two square roots as the square roots of the product. I rewrote the products in exponential form, it is easier.
  3. Now, if you remember, the square root of x means x at the exponent of 1/2: #sqrt(x)=x^(1/2)#.
    So in this case we can write:
    #sqrt(2^4)+sqrt(2^3)=2^(4/2)+2^(3/2)#
  4. Now we can solve the equation:
    #2^(4/2)+2^(3/2)=2^2+2^(1+1/2)=4+2*2^(1/2)=4+2sqrt(2)=2(2+sqrt(2))#.