InterviewSolution
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InterviewSolution
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Solve the following equations for `(x - 3)^(2)` and `(y + 2)^(2)` : `2x^(2) + y^(2) - 12x + 4y + 16 = 0` and `3x^(2) - 2y^(2) - 18x - 8y + 3 = 0`. |
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Answer» When it is said that solve for x and y then we have to find such values of x and y which satisfy both the equations. Similarly if we are asked that solve for `(x - 3)^(2)` and `(y - 2)^(2)`, then we have to find the value of `(x - 3)^(2)` and `(y + 2)^(2)` which satisfy the above two equations. First take the first equation as `2(x^(2) - 6x) + (y^(2) + 4y) = - 16` implies `2(x^(2)-6x+9) + (y^(2) + 4y + 4) = - 16 + 18 + 4` implies `2(x - 3)^(2) + (y + 2)^(2) = 6 " "....(1)` Now, take the second equation as `3(x^(2) - 6x) - 2 (y^(2) + 4y) = - 3` `implies 3(x^(2) - 6x + 9) - 2 (y^(2) + 4y + 4) = - 3 + 27 - 8` `implies 3(x - 3)^(2) - 2 (y-2)^(2) = 16" ....(2)"` Let `(x - 3)^(2)` = u and `(y + 2)^(2) = v`. So, equations (1) and (2) become 2u + v = 6 ....(3) and 3u - 2v = 16 ....(4) Multiplying equation (3) by 2, we get 4u + 2v = 12 ....(5) Adding equations (5) and (4), we get 4u + 2v = 12 `(3u - 2v = 16 )/(7u = 28)` `therefore u = 4 implies (x - 3)^(2) = 4` Putting u = 4 in equation (3), we get 2(4) + v = 6 implies v = - 2 implies `(y + 2)^(2) = - 2` But square of any real number cannot be negative. So, `(x - 3)^(2) = 4` and `(y + 2)^(2)` does not exist. |
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