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Solve `sqrt(3x^2-7x-30)+sqrt(2x^2-7x-5)=x+5.` |
Answer» Correct Answer - `x = 6, - 5//2` `sqrt(3x^(2) - 7x - 30) + sqrt(2x^(2) - 7x - 5) = x + 5` or `sqrt(3x^(2) - 7x - 30) =(x+5)- sqrt(2x^(2) - 7x - 5) = ` or On squaring, we get `3x^(2) - 7x - 30 =x^(2) + 10x + 25 - 2(x+5) xxsqrt(2x^(2) - 7x - 5)+ 2x^(2) - 7x - 5` or `10x + 50 = 2(x + 5) sqrt(2x^(2) - 7x - 5)` `rArr x = - 5 or sqrt(2x^(2) - 7x - 5) = 5` `rArr x = - 5 or 2x^(2) - 7x - 30 = 0` `rArr x = - 5 or x = 6 or x = - 5//2` But `x = - 5` does not satisfy the original equation. |
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