InterviewSolution
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InterviewSolution
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Prove that the following equations has no solutions. (i) `sqrt((2x+7))+sqrt((x+4))=0` (ii) `sqrt((x-4))=-5` (iii) `sqrt((6-x))-sqrt((x-8))=2` (iv) `sqrt(-2-x)=root(5)((x-7))` (v) `sqrt(x)+sqrt((x+16))=3` (vi) `7sqrt(x)+8sqrt(-x)+15/(x^(3))=98` (vii) `sqrt((x-3))-sqrt(x+9)=sqrt((x-1))` |
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Answer» (i) We have `sqrt((2x+7))+sqrt((x+4))=0` This equation is defined for `2x+7ge0` and `x+4ge0implies{(xge-7/2),(xge-4):"` `:.xge-7/2` For `xge-7/2` , the left hand side of the original equation is positive, but right hand side is zero. Therefore, the equation has no roots. (ii) We have `sqrt(x-4))=-5` The equation is defined for `x-4ge0` `:.xge4` For `xge4`, the left hand side of the original equation is positive but right hand side is negative. Therefore, the equation has no roots. (iii) We have `sqrt((5-x))-sqrt(x-8)=2` The equation is defined for `6-xge0` and `x-8ge0` `:.{(xle6),(xge8):}` Consequently, there is no `x` for which both expressions would have sense. Therefore, the equation has no roots. (iv) We have `sqrt((-2-x))=root(5)((x-7))` This equation is defined for `-2-xge0impliesxle-2` For `xle-2` the left hand side is positive, but right hand side is negative. Therefore, the equation has no roots. (v) We have `sqrt(x)+sqrt((x+16))=3` The equation is defined for `xge0` and `x+16ge0implies{(xge0),(xge-16):}` Hence `xge0` For `xge0` the left hand side `ge4`, but right hand side is 3. Therefore, the equation has no roots. (vi) We have `7sqrt(x)+8sqrt(-x)+15/(x^(3))=98` For `xlt0` the expression `7sqrt(x)` is meaningless, For `xgt0`, the expressionn `8sqrt(-x)` is meaningles and for `x=0` the expression `15/(x^(3))` is meaningless. Consequently the left hand side of the original equation ils meaningless for any `x epsilonR`. Therefore the equation has no roots. (vii) We have `sqrt((x-3))-sqrt((x+9))=sqrt(x-1)` This equation is defined for `{(x-3ge0),(x+9ge0),(x-1ge0):}implies{(xge3),(xge-9),(xge1):}` Hence `xge3` For `xge3, sqrt(x-3)ltsqrt(x+9)` i.e. `sqrt((x-3))-sqrt((x+9))lt0` Hence for `xge3`,the left hand side of the original equation is negative and right hand side is positive. Therefore, the equation has no roots. |
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