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Question Number 19643 by Tinkutara last updated on 13/Aug/17

$$\mathrm{Find}\mathrm{the}\mathrm{real}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}$$$$\sqrt{17+8x-2x^2}+\sqrt{4+12x-3x^2}=x^2$$$$-4x+13.$$

Answered by ajfour last updated on 13/Aug/17

$$\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}=\mathrm{u}-\mathrm{v}$$$$\Rightarrow \sqrt{\mathrm{u}}-\sqrt{\mathrm{v}}=1$$$$\sqrt{5^2-2{(2-\mathrm{x})}^2}=1+\sqrt{4^2-3{(2-\mathrm{x})}^2}$$$$\mathrm{let}{(2-\mathrm{x})}^2=\mathrm{t}$$$$\Rightarrow 25-2\mathrm{t}=1+16-3\mathrm{t}+2\sqrt{16-3\mathrm{t}}$$$$\Rightarrow {(\mathrm{t}+8)}^2=64-12\mathrm{t}$$$$\Rightarrow {\mathrm{t}}^2+28\mathrm{t}=0$$$$\mathrm{or}\mathrm{t}=0,-28;\mathrm{but}\mathrm{t}⩾0,\mathrm{so}$$$$\mathrm{t}=0,\mathrm{x}=2.$$$$\mathrm{At}\mathrm{x}=2,17+8\mathrm{x}-{2\mathrm{x}}^2>0$$$$\mathrm{and}4+12\mathrm{x}-{3\mathrm{x}}^2>0$$$$\mathrm{so}\mathrm{no}\mathrm{objection}\mathrm{to}\mathrm{x}=2.$$

Commented by aknabob last updated on 13/Aug/17

$$plsitsnotclear$$

Commented by Tinkutara last updated on 13/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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