x-x-x-x-175-8-

Question Number 140329 by benjo_mathlover last updated on 06/May/21

$$\:\mathrm{x}^{\lfloor\mathrm{x}\rfloor} \:+\:\mathrm{x}^{\lceil\mathrm{x}\rceil} \:=\:\frac{\mathrm{175}}{\mathrm{8}} \\ $$

Answered by john_santu last updated on 06/May/21

0<x<1→ x^0 + x^1 = ((175)/8) ⇒ x= ((167)/8)>20 1<x<2 ⇒x^1 + x^2 = ((175)/8) ⇒8x^2 +8x−175 =0 ⇒x = ((−8 ± (√(64+32.175)))/(16)) 2<x<3 ⇒x^2 +x^3 = ((175)/8) ⇒8x^3 +8x^2 −175=0 ⇒ (2x)^3 +2(2x)^2 −175=0 let 2x = y ⇒y^3 +2y^2 −175=0 (y−5)(y^2 +5y+35)=0 ⇒y=5=2x ; x = (5/2). (solution)

$$\:\mathrm{0}<{x}<\mathrm{1}\rightarrow\:{x}^{\mathrm{0}} \:+\:{x}^{\mathrm{1}} \:=\:\frac{\mathrm{175}}{\mathrm{8}} \\ $$$$\Rightarrow\:{x}=\:\frac{\mathrm{167}}{\mathrm{8}}>\mathrm{20} \\ $$$$\:\mathrm{1}<{x}<\mathrm{2}\:\Rightarrow{x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:=\:\frac{\mathrm{175}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{8}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{175}\:=\mathrm{0} \\ $$$$\Rightarrow{x}\:=\:\frac{−\mathrm{8}\:\pm\:\sqrt{\mathrm{64}+\mathrm{32}.\mathrm{175}}}{\mathrm{16}} \\ $$$$\mathrm{2}<{x}<\mathrm{3}\:\Rightarrow{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \:=\:\frac{\mathrm{175}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{8}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{175}=\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{2}{x}\right)^{\mathrm{3}} +\mathrm{2}\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{175}=\mathrm{0} \\ $$$${let}\:\mathrm{2}{x}\:=\:{y}\:\Rightarrow{y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{175}=\mathrm{0} \\ $$$$\left({y}−\mathrm{5}\right)\left({y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{35}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{5}=\mathrm{2}{x}\:;\:{x}\:=\:\frac{\mathrm{5}}{\mathrm{2}}.\:\left({solution}\right) \\ $$

Answered by mr W last updated on 06/May/21

let x=n+f with 0≤f<1 (n+f)^n +(n+f)^(n+1) =((175)/8) ((175)/8)=(n+f)^n +(n+f)^(n+1) ≥n^n +n^(n+1) =(n+1)n^n ⇒n≤2 ((175)/8)=(n+f)^n +(n+f)^(n+1) ≤(n+1)^n +(n+1)^(n+1) =(n+2)(n+1)^n ⇒n≥2 ⇒n=2 x^2 +x^3 =((175)/8) (x−(5/2))(x^2 +((7x)/2)+((35)/4))=0 ⇒x=(5/2)

$${let}\:{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\left({n}+{f}\right)^{{n}} +\left({n}+{f}\right)^{{n}+\mathrm{1}} =\frac{\mathrm{175}}{\mathrm{8}} \\ $$$$\frac{\mathrm{175}}{\mathrm{8}}=\left({n}+{f}\right)^{{n}} +\left({n}+{f}\right)^{{n}+\mathrm{1}} \geqslant{n}^{{n}} +{n}^{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){n}^{{n}} \\ $$$$\Rightarrow{n}\leqslant\mathrm{2} \\ $$$$\frac{\mathrm{175}}{\mathrm{8}}=\left({n}+{f}\right)^{{n}} +\left({n}+{f}\right)^{{n}+\mathrm{1}} \leqslant\left({n}+\mathrm{1}\right)^{{n}} +\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} =\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)^{{n}} \\ $$$$\Rightarrow{n}\geqslant\mathrm{2} \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{3}} =\frac{\mathrm{175}}{\mathrm{8}} \\ $$$$\left({x}−\frac{\mathrm{5}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{7}{x}}{\mathrm{2}}+\frac{\mathrm{35}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

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