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

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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}$Case $1$: $x + 5 = 0 \Rightarrow x = -5$. Checking in the original equation: $\sqrt{75+35-30} + \sqrt{50+35-5} = 0 \Rightarrow \sqrt{80} + \sqrt{80} = 0$, which is false.Case $2$: $5 = \sqrt {2x^2 - 7x - 5}$Squaring both sides: $25 = 2x^2 - 7x - 5$$2x^2 - 7x - 30 = 0$$(2x + 5)(x - 6) = 0$$x = 6$ or $x = -2.5$. Checking $x = 6$ in the original equation: $\sqrt{108-42-30} + \sqrt{72-42-5} = 6+5 \Rightarrow \sqrt{36} + \sqrt{25} = 11 \Rightarrow 6+5 = 11$. This is true.Thus, $x = 6$.

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

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