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Question Number 19409 by Tinkutara last updated on 10/Aug/17

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{of}\mathrm{the}$$$$\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}{x}^2-7\left[x\right]+5=0?$$$$(\mathrm{Here}\left[x\right]\mathrm{denotes}\mathrm{the}\mathrm{greatest}\mathrm{integer}$$$$\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}x.\mathrm{For}\mathrm{example}$$$$[3.4]=3\mathrm{and}[-2.3]=-3.)$$

Commented by mrW1 last updated on 11/Aug/17

$$2+23+30+37=92$$$$\mathrm{is}\mathrm{this}\mathrm{right}?$$

Answered by mrW1 last updated on 11/Aug/17

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-7\left[\mathrm{x}\right]+5$$$$\mathrm{let}\mathrm{x}=\mathrm{f}+\mathrm{d}\mathrm{with}0⩽\mathrm{d}<1$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={(\mathrm{f}+\mathrm{d})}^2-7\mathrm{f}+5={\mathrm{f}}^2+(2\mathrm{d}-7)\mathrm{f}+({\mathrm{d}}^2+5)$$$$\mathrm{f}\left(\mathrm{x}\right)⩾{\mathrm{f}}^2-7\mathrm{f}+5=\mathrm{g}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)<{\mathrm{f}}^2-5\mathrm{f}+6=\mathrm{h}\left(\mathrm{x}\right)$$$$\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^2-7\mathrm{f}+5=(\mathrm{f}-\frac{7-\sqrt{29}}{2})(\mathrm{f}-\frac{7+\sqrt{29}}{2})$$$$\mathrm{h}\left(\mathrm{x}\right)={\mathrm{f}}^2-5\mathrm{f}+6=(\mathrm{f}-2)(\mathrm{f}-3)$$$$\mathrm{since}\mathrm{g}\left(\mathrm{x}\right)⩽\mathrm{f}\left(\mathrm{x}\right)<\mathrm{h}\left(\mathrm{x}\right),$$$$\mathrm{solution}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)=0:$$$$\frac{7-\sqrt{29}}{2}\approx 0.807⩽\mathrm{f}<2\mathrm{or}$$$$3<\mathrm{f}⩽\frac{7+\sqrt{29}}{2}\approx 6.192$$$$\Rightarrow \mathrm{f}=1,4,5,6$$$$$$$$\mathrm{with}\mathrm{x}=1+\mathrm{d}:$$$${(1+\mathrm{d})}^2-7\times 1+5=0$$$$\Rightarrow \mathrm{d}=\sqrt{2}-1$$$$\Rightarrow \mathrm{x}=1+\sqrt{2}-1=\sqrt{2}$$$$\mathrm{with}\mathrm{x}=4+\mathrm{d}:$$$${(4+\mathrm{d})}^2-7\times 4+5=0$$$$\Rightarrow \mathrm{d}=\sqrt{23}-4$$$$\Rightarrow \mathrm{x}=4+\sqrt{23}-4=\sqrt{23}$$$$\mathrm{with}\mathrm{x}=5+\mathrm{d}:$$$${(5+\mathrm{d})}^2-7\times 5+5=0$$$$\Rightarrow \mathrm{d}=\sqrt{30}-5$$$$\Rightarrow \mathrm{x}=5+\sqrt{30}-5=\sqrt{30}$$$$\mathrm{with}\mathrm{x}=6+\mathrm{d}:$$$${(6+\mathrm{d})}^2-7\times 6+5=0$$$$\Rightarrow \mathrm{d}=\sqrt{37}-6$$$$\Rightarrow \mathrm{x}=6+\sqrt{37}-6=\sqrt{37}$$$$$$$$\mathrm{all}\mathrm{solutions}:$$$$\sqrt{2},\sqrt{23},\sqrt{30},\sqrt{37}$$$$\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{squares}:$$$$2+23+30+37=92$$

Commented by mrW1 last updated on 11/Aug/17

Commented by Tinkutara last updated on 12/Aug/17

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

Answered by ajfour last updated on 12/Aug/17

$$\mathrm{let}\mathrm{x}=\mathrm{n}+\mathrm{f}$$$$\mathrm{let}\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\}=\mathrm{n}+\mathrm{f}$$$$\mathrm{then}{(\mathrm{n}+\mathrm{f})}^2+5-7\mathrm{n}=0$$$$\Rightarrow {(\mathrm{n}+\mathrm{f})}^2+5\mathrm{is}\mathrm{an}\mathrm{integer}=7\mathrm{n}$$$$\mathrm{clearly}0<\mathrm{n}<7\mathrm{and}0⩽\mathrm{f}<1$$$$\mathrm{x}=\mathrm{n}+\mathrm{f}=\sqrt{7\mathrm{n}-5},\mathrm{f}=\mathrm{x}-\mathrm{n}$$$$\mathrm{for}\mathrm{n}=1,\mathrm{x}=\sqrt{2},\mathrm{f}=\sqrt{2}-1$$$$\mathrm{n}=2,\mathrm{x}=3,\mathrm{f}=1$$$$\mathrm{so}\mathrm{not}\mathrm{acceptable};$$$$\mathrm{n}=3,\mathrm{x}=4,\mathrm{f}=1$$$$\mathrm{not}\mathrm{acceptable};$$$$\mathrm{n}=4,\mathrm{x}=\sqrt{23},\mathrm{f}=\sqrt{23}-4$$$$\mathrm{this}\mathrm{is}\mathrm{quite}\mathrm{alright}!$$$$\mathrm{n}=5,\mathrm{x}=\sqrt{30},\mathrm{f}=\sqrt{30}-5$$$$\mathrm{and}\mathrm{if}\mathrm{n}=6,\mathrm{x}=\sqrt{37},\mathrm{f}=\sqrt{37}-6$$$$\mathrm{acceptable}\mathrm{then}\mathrm{are}\mathrm{n}=1,4,5,6$$$$\mathrm{\Sigma }{\mathrm{x}}^2=\mathrm{\Sigma }(7\mathrm{n}-5)$$$$=2+23+30+37$$$$=92.$$$$$$

Commented by ajfour last updated on 12/Aug/17

$$\mathrm{as}{(\mathrm{n}+\mathrm{f})}^2+5=7\mathrm{n}$$$$\mathrm{L}.\mathrm{H}.\mathrm{S}.⩾5\Rightarrow 7\mathrm{n}⩾5\mathrm{so}\mathrm{n}>0$$$$\mathrm{for}\mathrm{n}>7,{\mathrm{n}}^2>7\mathrm{n}$$$$\Rightarrow {(\mathrm{n}+\mathrm{f})}^2+5>7\mathrm{n}$$$$\mathrm{so}\mathrm{n}<7\mathrm{if}{(\mathrm{n}+\mathrm{f})}^2+5=7\mathrm{n}.$$

Commented by Tinkutara last updated on 12/Aug/17

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

Commented by ajfour last updated on 12/Aug/17

Commented by ajfour last updated on 12/Aug/17

$$\mathrm{y}=7\mathrm{x};\mathrm{y}={\mathrm{x}}^2+5;\mathrm{y}=7\left[\mathrm{x}\right]$$$$$$

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