#asqrt(a)+bsqrt(b)=183# and #asqrt(b)+bsqrt(a)=182# Find #9/5(a+b)# ?

2 Answers
Apr 5, 2018

73

Explanation:

Let #sqrta =x => x^2=a#
and #sqrtb =y => y^2=b#

So, the altered equations now look like,
#x^ 3 +y^ 3 =183# and #x^2y+y^ 2x=182#
#color(white)(wwwwwwwwwd# #=>xy(x+y) = 182#

We know the basic identity,
#color(magenta)((x+y)^3=x^ 3 +y^ 3 +3(x^ 2 y+y^ 2 x)#

Plugging in the "altered" equations,
#=>(x+y)^3=183 +3(182)#
#=>(x+y)^3=729#
#=> color(blue)(x+y = 9#

Now, plug #x+y = 9# into #xy(x+y) = 182#
#=>color(red)( xy = 182/9#

Also, #(x-y)^2 = (color(blue)(x+y))^2 - 4color(red)(xy)#

#=> color(blue)(9)^2 - 4(color(red)(182/9))#

#=> 81-(4(182))/9 = 1/9#

#=> x-y = +-1/3#

Since we've found out #x+y = 9# and #x-y = +-1/3#, we can find the values of #x# and #y# by elimination method.

When #x-y = + 1/3# #color(white)(wwwwwwwwwd# When #x-y = - 1/3#

Adding both the equations, #color(white)(wwwwwd# Adding both the equations,

#2x = 9+1/3# #color(white)(wwwwwwwwwwwwwwd# #2x = 9-1/3#

#x= 14/3# #color(white)(wwwwwwwwwwwwwwwwww##x= 13/3#

therefore, #color(white)(wwwwwwwwwwwwwwwwd# therefore,

#y=13/3# #color(white)(wwwwwwwwwwwwwwwwwd# #y=14/3#

Either way, we have to find, #9/5 (a+b) => 9/5 (x^2 + y^2)#

#=> 9/5 ((13/3)^2 + (14/3)^2)#

#=> cancel9/5 xx (169 + 196) /cancel9#

#=>73#

Apr 5, 2018

Alternate to the previous answer

Explanation:

Let #sqrta =x => x^2=a#
and #sqrtb =y => y^2=b#

So, the altered equations now look like,
#x^ 3 +y^ 3 =183# and #x^2y+y^ 2x=182#
#color(white)(wwwwwwwwwd# #=>xy(x+y) = 182#

We know the basic identity,
#color(magenta)((x+y)^3=x^ 3 +y^ 3 +3(x^ 2 y+y^ 2 x)#

Plugging in the "altered" equations,
#=>(x+y)^3=183 +3(182)#
#=>(x+y)^3=729#
#=> color(blue)(x+y = 9#

Now, plug #x+y = 9# into #xy(x+y) = 182#
#=>color(red)( xy = 182/9#

We know, #(x+y)^2 = x^2 + y^2 +2xy#
#=> x^2 + y^2 = (x+y)^2 - 2xy#
#color(white)(wwwwwwi# #= 9^2 - 2(182/9)#

#color(white)(wwwwwwi# #= 365/9#

We need to find, #9/5 (a+b) => 9/5 (x^2 + y^2)#

#=> 9/5 (365/9)#

#=> cancel9/5 xx (365) /cancel9#

#=>73#

#color(white)(wwwwwwi#

#color(white)(wwwwwwi#

#color(white)(wwwwwwi#

Another way could be,
since,
#x³+y³=183#

#(x+y)(x²+y²−xy)=183#

#(x+y)(x²+y²)−xy(x+y)=183#

#(x+y)(x²+y²)=182+183 =365#

#(x²+y²)=365/9#
.
Then the required expression #= (9/5)(a+b)=(9/5)(x²+y²)#
#=(9/5)(3659)=73 #