Solve `sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39.)`

Answer

`x = 3`

Explanation

We have
`sqrt(x^(2) + 4x - 21) + sqrt(x^(2) - x - 6) = sqrt(6x^(2) - 5x - 39)`
or `sqrt((x + 7)(x - 3)) + sqrt((x - 3)(x - 2)) = sqrt((x-3)((6x + 13))) (1)`
or ` sqrt(x - 3) + (sqrt(x - 7) = sqrt(x-2)-sqrt(6x + 13) ) `
`sqrt(x - 3) =0 or (sqrt(x - 7) = sqrt(x+2)-sqrt(6x + 13)= 0 `
`rArr x = 3 - or sqrt(x + 7) + sqrt(x + 2) = sqrt(6x = 13) `
Now , `sqrt(x + 7) + sqrt(x + 2) = sqrt(6x + 13)`
or `(sqrt(x + 7) + sqrt(x + 2))^(2) = sqrt(6x + 13)`
or `
or `x + 7 + x + 2+2 = sqrt((x - 7) = (x-2)) = 6x + 13`
or `2x + 9+2 = sqrt((x - 7) = (x-2)) = 6x + 13`
or `2sqrt((x - 7) = (x-2)) = 4x+ 4`
`sqrt((x - 7) = (x-2)) = 2(x+1)`
or `(x - 7) = (x-2) = 4(x+1)^(2)` (squraring both sides)
or `x^(2) + 9x + 14 = 4 (x^(2)+ 2x + 1)`
or ` 3x^(2) - x - 10 = 0`
or `(x - 2) (3x + 5) = 0`
`rArr x = 2 or x = (-5)/(3) ,` which does not satisfy (1).
Hence , x = 3 only.