Find z (floor). Please? Add work and explanation.

Question

computer

748 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-14T14:09:42

$z=1/5(2sqrt2+sqrt3)$

$z= (sqrt3^2 - 2sqrt2^2)/(sqrt3-2sqrt2)$

$= (3-2xx2)/(sqrt3-2sqrt2)$

$= (-1)/(sqrt3-2sqrt2)$

Rationalise the denominator:

$= ((-1)xx(sqrt3+2sqrt2))/((sqrt3-2sqrt2)xx(sqrt3+2sqrt2))$

$= (-(sqrt3+2sqrt2))/(3-4xx2)$

$= (-(sqrt3+2sqrt2))/(3-8)$

$= (sqrt3+2sqrt2)/5$

$=1/5(2sqrt2+sqrt3)$