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Solve `sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39.)` |
Answer» Correct Answer - `x = 3` 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. |
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