How do you simplify #sqrt(2/3)+sqrt(3/2)#?

Question

Isaiah Anthony · Accepted Answer

#(5sqrt(6))/6#

From the given expression #sqrt(2/3)+sqrt(3/2)#

#sqrt(2/3)+sqrt(3/2)#

#sqrt(2/3*3/3)+sqrt(3/2*2/2)#

#sqrt(6/9)+sqrt(6/4)#

#sqrt(6)/3*2/2+sqrt(6)/2*3/3#

#(2sqrt(6))/6+(3*sqrt(6))/6#

#(5sqrt(6))/6#

For what I know , radicals are already on its simplest form when there is no radical at the denominator, there is no fraction in the radicand, and the radicand is not reducible anymore.

God bless...I hope the explanation is useful.

Jace Anderson · Answer

undefined To simplify sqrt(2/3) + sqrt(3/2), we can rationalize the denominators by multiplying the numerator and denominator of each square root by the conjugate of the denominator.

For sqrt(2/3), we multiply the numerator and denominator by sqrt(3), resulting in (sqrt(2) * sqrt(3)) / (sqrt(3) * sqrt(3)), which simplifies to sqrt(6) / 3.

For sqrt(3/2), we multiply the numerator and denominator by sqrt(2), giving us (sqrt(3) * sqrt(2)) / (sqrt(2) * sqrt(2)), which simplifies to sqrt(6) / 2.

Combining the simplified expressions, we have sqrt(6) / 3 + sqrt(6) / 2. To add these fractions, we need a common denominator, which is 6.

Converting the fractions to have a denominator of 6, we get (2 * sqrt(6)) / 6 + (3 * sqrt(6)) / 6.

Adding the numerators, we have (2 * sqrt(6) + 3 * sqrt(6)) / 6, which simplifies to (5 * sqrt(6)) / 6.

Therefore, sqrt(2/3) + sqrt(3/2) simplifies to (5 * sqrt(6)) / 6.