Question

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

Answer 1

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`

Answer 2

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`