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Question Number 111934 by mohammad17 last updated on 05/Sep/20

Answered by Aina Samuel Temidayo last updated on 05/Sep/20

$$\sqrt{\mathrm{x}+5}+\sqrt{2\mathrm{x}+1}=4\mathrm{x}-10$$$$\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{have}$$$$\mathrm{x}+5+2\mathrm{x}+1+2\sqrt{(\mathrm{x}+5)(2\mathrm{x}+1)}={16\mathrm{x}}^2+100-80\mathrm{x}$$$$\Rightarrow$$$${16\mathrm{x}}^2-80\mathrm{x}-3\mathrm{x}+100-6=2\sqrt{(\mathrm{x}+5)(2\mathrm{x}+1)}$$$$\Rightarrow$$$${({16\mathrm{x}}^2-83\mathrm{x}+94)}^2={\left(2\sqrt{(\mathrm{x}+5)(2\mathrm{x}+1)}\right)}^2$$$$\Rightarrow {256\mathrm{x}}^4-{2656\mathrm{x}}^3+{9897\mathrm{x}}^2-15604\mathrm{x}+8836$$$$=4({2\mathrm{x}}^2+11\mathrm{x}+5)$$$$\Rightarrow {256\mathrm{x}}^4-{2656\mathrm{x}}^3+{9897\mathrm{x}}^2-{8\mathrm{x}}^2-15604\mathrm{x}-44\mathrm{x}+8836-20=0$$$$\Rightarrow (\mathrm{x}-4)({256\mathrm{x}}^3-{1632\mathrm{x}}^2+3361\mathrm{x}-2204)=0$$$$\Rightarrow \mathrm{x}=4\mathrm{or}$$$${256\mathrm{x}}^3-{1632\mathrm{x}}^2+3361\mathrm{x}-2204=0$$$$$$$$\mathrm{Factors}\mathrm{of}2204:$$$$\pm 1,\pm 2,\pm 19,\pm 29,\pm 38,\pm 58,\pm 76,\pm 116,\pm 551,\pm 1102,\pm 2204.$$$$\mathrm{Substituting}\mathrm{any}\mathrm{of}\mathrm{these}\mathrm{in}$$$${256\mathrm{x}}^3-{1632\mathrm{x}}^2+3361\mathrm{x}-2204{\mathrm{won}}^{\prime }\mathrm{t}$$$$\mathrm{give}\mathrm{us}0\iff \mathrm{there}\mathrm{are}\mathrm{no}$$$$\mathrm{integer}\mathrm{solutions}\mathrm{to}\mathrm{it}.$$$$\Rightarrow 4\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{integer}\mathrm{solution}.$$$$\Rightarrow \mathrm{x}=4$$$$$$$${\mathrm{x}}^{{\mathrm{x}}^2-3\mathrm{x}}=4^{{\left(4\right)}^2-3\left(4\right)}=4^{16-12}=4^4=256.$$

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