If sqrt{3x^2 - 7x - 30} + sqrt{2x^2 - 7x - 5} | vedclass

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(C) Given equation: $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$Rearranging the terms: $\sqrt{3x^2 - 7x - 30} = (x + 5) - \sqrt{2x^2 - 7x -

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$6$

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(C) Given equation: $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$Rearranging the terms: $\sqrt{3x^2 - 7x - 30} = (x + 5) - \sqrt{2x^2 - 7x - 5}$Squaring both sides:$3x^2 - 7x - 30 = (x + 5)^2 + (2x^2 - 7x - 5) - 2(x + 5)\sqrt{2x^2 - 7x - 5}$$3x^2 - 7x - 30 = x^2 + 10x + 25 + 2x^2 - 7x - 5 - 2(x + 5)\sqrt{2x^2 - 7x - 5}$$3x^2 - 7x - 30 = 3x^2 + 3x + 20 - 2(x + 5)\sqrt{2x^2 - 7x - 5}$$-10x - 50 = -2(x + 5)\sqrt{2x^2 - 7x - 5}$$5(x + 5) = (x + 5)\sqrt{2x^2 - 7x - 5}$This implies either $x + 5 = 0$ or $5 = \sqrt{2x^2 - 7x - 5}$.If $x = -5$,the original equation becomes $\sqrt{75 + 35 - 30} + \sqrt{50 + 35 - 5} = 0$,which is $\sqrt{80} + \sqrt{80} = 0$,which is false.If $5 = \sqrt{2x^2 - 7x - 5}$,then $25 = 2x^2 - 7x - 5$,which gives $2x^2 - 7x - 30 = 0$.Factoring the quadratic: $(2x + 5)(x - 6) = 0$.Thus,$x = 6$ or $x = -2.5$. Checking $x = 6$ in the original equation: $\sqrt{108 - 42 - 30} + \sqrt{72 - 42 - 5} = \sqrt{36} + \sqrt{25} = 6 + 5 = 11$,which matches $x + 5 = 11$. Therefore,$x = 6$ is the solution.

If $\sqrt{3x^2 - 7x - 30} + \sqrt{2x^2 - 7x - 5} = x + 5$,then $x$ is equal to

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