If #(sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3)) = a+sqrt(15)b# for integers #a# and #b# then what are #a# and #b# ?

Question

Julia Adams · Accepted Answer

#a=4"; "b=1#

Premise: The appropriate query ought to be:

#(sqrt5+sqrt3)/(sqrt5-sqrt3)=a+sqrt(15)color(white)(.) b#

#color(blue)("Consider the left hand side (LHS) only")#

#color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx1) )#

#color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx (sqrt5+sqrt3)/(sqrt5+sqrt3) )#

#((sqrt5+sqrt3)(sqrt5+sqrt3))/(5-3)#

#(5+2sqrt(3xx5)+3)/2#

#8/2+(2sqrt(15))/2#

#4+sqrt15# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Putting it all together")#

#4+sqrt15=a+sqrt15color(white)(..)b#

Thus #a=4"; "b=1# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Additional note")#

If the RHS #->a+sqrt(15bcolor(white)(..))#

The #b# is still 1

Savannah Anderson · Answer

#a=4" "# and #" "b=1# To rational the denominator #sqrt(5)-sqrt(3)# we can multiply both numerator and denominator by the radical conjugate #sqrt(5)+sqrt(3)#... #(sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3)) = ((sqrt(5)+sqrt(3))(sqrt(5)+sqrt(3)))/((sqrt(5)-sqrt(3))(sqrt(5)+sqrt(3)))# #color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = ((sqrt(5))^2+2sqrt(5)sqrt(3)+(sqrt(3))^2)/((sqrt(5))^2-(sqrt(3))^2)# #color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (5+2sqrt(15)+3)/(5-3)# #color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (8+2sqrt(15))/2# #color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = 4+sqrt(15)# So #a=4# and #b=1#

Isaiah Anthony · Answer

undefined a = 2, b = 1