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Solving the following inequalities: $$\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$$

I have been solving some questions on inequalities lately and I have come across this question, but I can't figure out a way to solve this...it contains both log and square root. Here's how I approached the question:

$\sqrt{x^2-7x+10}+9\log_4{\frac{x}{8}}\geq 2x+\sqrt{14x-20-2x^2}-13$

$\sqrt{x^2-7x+10}+9\log_4{x}-3/2\geq 2x+\sqrt{(-2)(x^2-7x+10)}-13$

$(1-2i)\sqrt{x^2-7x+10}+\geq 2x-9\log_4{x}+3/2-13$

$(1-2i)\sqrt{(x-5)(x-2)}+\geq 2x-9\log_4{x}-23/2$

I don't know how to proceed further. It would be great if I could get a hint.

2 Answers 2

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Hint:

Let $x \in \mathbb R$

If $$\sqrt{x^2-7x+10}=\sqrt{(x-2)(x-5)}$$ then $x\le 2$ or$x \ge5$

If $$\sqrt{-2(x^2-7x+10)}=\sqrt{-2(x-2)(x-5)}$$ then $2\le x \le 5$

Hence, $x=2$ or $x=5$

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    Oh! That seems pretty simple...thank you! – 2017-01-11
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    This is just necessarily condition. – 2017-01-12
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If $x \in \mathbb R $, then we have that the expressions $\sqrt {x^2-7x-10} \text { and } \sqrt {20+14x-2x^2}$ should be positive, that is, $\geq 0$. But one expression is $\sqrt {2}i $ times the other implying that both expressions are zero, that is, $x=2$ and $x=5$ can satisfy this inequality.

But however $x=5$ does not satisfy the inequality giving us $\boxed {x=2} $ is the only solution. Hope it helps.

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    Sorry @Roman83 I didn't load your answer while I was writing mine. So, my +1 to you. – 2017-01-11
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    my +1 to you too – 2017-01-11