How do you simplify #sqrt(5.6)#?

1 Answer
Apr 27, 2018

Answer:

#(2sqrt35)/5=2/5sqrt35#

Explanation:

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

#•color(white)(x)sqrtaxxsqrtb=sqrt(ab)" and "sqrtaxxsqrta=a#

#•color(white)(x)sqrt(a/b)=sqrta/sqrtb#

#"note that "5.6=56/10#

#rArrsqrt5.6=sqrt(56/10)=sqrt56/sqrt10#

#"and "sqrt56=sqrt(4xx14)=sqrt4xxsqrt14=2sqrt14#

#rArrsqrt56/sqrt10=(2sqrt14)/sqrt10#

#color(blue)"rationalise the denominator"#

#"multiply numerator/denominator by "sqrt10#

#rArr(2sqrt14)/sqrt10xxsqrt10/sqrt10#

#=(2sqrt140)/10=(2xx2sqrt35)/10=(4sqrt35)/10=2/5sqrt35#