Question-224342

Question Number 224342 by Rojarani last updated on 04/Sep/25

Answered by mingski last updated on 04/Sep/25

x^2 −x−1000(√(1+8000x))=1000 x^2 −x=1000(1+(√(1+8000x))) x(x−1)=1000(1+(√(1+8000x))) Then guess the value of x. x=1000k or 1000k+1 we found x=2001.

$${x}^{\mathrm{2}} −{x}−\mathrm{1000}\sqrt{\mathrm{1}+\mathrm{8000}{x}}=\mathrm{1000} \\ $$$${x}^{\mathrm{2}} −{x}=\mathrm{1000}\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8000}{x}}\right) \\ $$$${x}\left({x}−\mathrm{1}\right)=\mathrm{1000}\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8000}{x}}\right) \\ $$$$\mathrm{Then}\:\mathrm{guess}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$${x}=\mathrm{1000}{k}\:{or}\:\mathrm{1000}{k}+\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{found}\:{x}=\mathrm{2001}. \\ $$

Commented by Frix last updated on 04/Sep/25

Since this is a math forum we don′t guess.

$$\mathrm{Since}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{math}\:\mathrm{forum}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{guess}. \\ $$

Answered by Rasheed.Sindhi last updated on 04/Sep/25

(√(1+8000x)) =y x=((y^2 −1)/(8000)) (((y^2 −1)/(8000)))^2 −((y^2 −1)/(8000))−1000y−1000=0 (((y^2 −1)/(8k)))^2 −((y^2 −1)/(8k))−ky−k=0; k=1000 ((y^4 −2y^2 +1)/(64k^2 ))−((y^2 −1)/(8k))−ky−k=0 y^4 −2y^2 +1−8ky^2 +8k−8k^2 y−8k^2 =0 y^4 −2y^2 −8ky^2 −8k^2 y−8k^2 +8k+1=0 y^4 −2(1+4k)y^2 −8k^2 y−8k^2 +8k+1=0 ...

$$\sqrt{\mathrm{1}+\mathrm{8000}{x}}\:={y} \\ $$$${x}=\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8000}} \\ $$$$\left(\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8000}}\right)^{\mathrm{2}} −\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8000}}−\mathrm{1000}{y}−\mathrm{1000}=\mathrm{0} \\ $$$$\left(\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8}{k}}\right)^{\mathrm{2}} −\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8}{k}}−{ky}−{k}=\mathrm{0};\:\:{k}=\mathrm{1000} \\ $$$$\frac{{y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{64}{k}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} −\mathrm{1}}{\mathrm{8}{k}}−{ky}−{k}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{8}{ky}^{\mathrm{2}} +\mathrm{8}{k}−\mathrm{8}{k}^{\mathrm{2}} {y}−\mathrm{8}{k}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} −\mathrm{8}{ky}^{\mathrm{2}} −\mathrm{8}{k}^{\mathrm{2}} {y}−\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{2}\left(\mathrm{1}+\mathrm{4}{k}\right){y}^{\mathrm{2}} −\mathrm{8}{k}^{\mathrm{2}} {y}−\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{1}=\mathrm{0} \\ $$$$… \\ $$

Answered by Frix last updated on 04/Sep/25

x^2 −x−p=p(√(8px+1)) Squaring & transforming x(x^3 −2x^2 −(2p−1)x−2p(2p−1)(2p+1))=0 Obviously x≠0 x^3 −2x^2 −(2p−1)x−2p(2p−1)(2p+1)=0 First, try the factors of the constant ⇒ x=2p+1 [this is always a solution for p≥−(1/4)] Test: (2p+1)^2 −(2p+1)−p=p(√(8p(2p+1)+1)) p(4p+1)=p∣4p+1∣ true for p≥−(1/4)

$${x}^{\mathrm{2}} −{x}−{p}={p}\sqrt{\mathrm{8}{px}+\mathrm{1}} \\ $$$$\mathrm{Squaring}\:\&\:\mathrm{transforming} \\ $$$${x}\left({x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\left(\mathrm{2}{p}−\mathrm{1}\right){x}−\mathrm{2}{p}\left(\mathrm{2}{p}−\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$$$\mathrm{Obviously}\:{x}\neq\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\left(\mathrm{2}{p}−\mathrm{1}\right){x}−\mathrm{2}{p}\left(\mathrm{2}{p}−\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{First},\:\mathrm{try}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant} \\ $$$$\Rightarrow\:{x}=\mathrm{2}{p}+\mathrm{1} \\ $$$$\left[\mathrm{this}\:\mathrm{is}\:\mathrm{always}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{for}\:{p}\geqslant−\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$$$ \\ $$$$\mathrm{Test}: \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2}{p}+\mathrm{1}\right)−{p}={p}\sqrt{\mathrm{8}{p}\left(\mathrm{2}{p}+\mathrm{1}\right)+\mathrm{1}} \\ $$$${p}\left(\mathrm{4}{p}+\mathrm{1}\right)={p}\mid\mathrm{4}{p}+\mathrm{1}\mid \\ $$$$\mathrm{true}\:\mathrm{for}\:{p}\geqslant−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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