How do you solve #\sqrt { 8} \cdot \sqrt { 18} + \sqrt { 50} \cdot \sqrt { 2}#?

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Answer 1 · 2017-11-13T12:55:06

#22#

begin by rewriting the square roots in theirs simplest form

#sqrt8*sqrt18+sqrt50*sqrt2#

#=sqrt(2xx4)*sqrt(9xx2)+sqrt(25xx2)*sqrt2#

#=>2sqrt2*3sqrt2+5sqrt2sqrt2#

#=6xxsqrt2xxsqrt2+5sqrt2xxsqrt2#

#=6xx2+5xx2#

#=12+10=22#

Answer 2 · 2017-11-13T13:52:57

See explanation.

We can use the following rule to calculate the value of this expression:

The product of square roots of expressions is equal to square root of product of those experssions

#sqrt(a)sqrt(b)=sqrt(ab)#

Here we get:

#sqrt(8)*sqrt(18)+sqrt(50)*sqrt(2)=sqrt(8*18)+sqrt(50*2)=#

#=sqrt(144)+sqrt(100)=12+10=22#