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Question Number 162043 by HongKing last updated on 25/Dec/21

Find valu of x if x∈R ((9x - 1))^(1/3) + (√(8x - 1)) + ((8x + 15))^(1/4) - (5/2) = 0

$$\mathrm{Find}\:\mathrm{valu}\:\mathrm{of}\:\:\boldsymbol{\mathrm{x}}\:\:\mathrm{if}\:\:\mathrm{x}\in\mathbb{R}\: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{9x}\:-\:\mathrm{1}}\:+\:\sqrt{\mathrm{8x}\:-\:\mathrm{1}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{8x}\:+\:\mathrm{15}}\:-\:\frac{\mathrm{5}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$

Commented by mr W last updated on 25/Dec/21

with x=(1/8) (((9/8)−1))^(1/3) +(√((8/8)−1))+(((8/8)+15))^(1/4) −(5/2) =((1/8))^(1/3) +(√0)+((16))^(1/4) −(5/2) =(1/2)+0+2−(5/2) =0 ✓ ⇒x=(1/8) is one and the only one root, since f(x) is strictly increasing.

$${with}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{8}}−\mathrm{1}}+\sqrt{\frac{\mathrm{8}}{\mathrm{8}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{15}}−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{8}}}+\sqrt{\mathrm{0}}+\sqrt[{\mathrm{4}}]{\mathrm{16}}−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{0}+\mathrm{2}−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$=\mathrm{0}\:\checkmark \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{8}}\:{is}\:{one}\:{and}\:{the}\:{only}\:{one}\:{root}, \\ $$$${since}\:{f}\left({x}\right)\:{is}\:{strictly}\:{increasing}. \\ $$

Commented by HongKing last updated on 26/Dec/21

thank you my dear Sir cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

Answered by aleks041103 last updated on 25/Dec/21

f(x)=((9x−1))^(1/3) +(√(8x−1))+((8x+15))^(1/4) by observation f(x) is monotonously rising ⇒there is at most 1 solution to f(x)=(5/2) f(x)=(5/2) ⇒((72x−8))^(1/3) +(√(32x−4))+((128x+240))^(1/4) =5 for f(x)∈R⇒ 32x−4≥0⇒x≥(1/8) 128x+240≥0⇒x≥−((240)/(128)) ⇒x≥(1/8) try x=(1/8) ⇒((72x−8))^(1/3) +(√(32x−4))+((128x+240))^(1/4) = =((9−8))^(1/3) +(√(4−4))+((16+240))^(1/4) = =1+0+4=5 ⇒x=(1/8) is a solution ⇒the only solution of f(x)=(5/2) over R is x=(1/8)

$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{9}{x}−\mathrm{1}}+\sqrt{\mathrm{8}{x}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\mathrm{8}{x}+\mathrm{15}} \\ $$$${by}\:{observation}\:{f}\left({x}\right)\:{is}\:{monotonously}\:{rising} \\ $$$$\Rightarrow{there}\:{is}\:{at}\:{most}\:\mathrm{1}\:{solution}\:{to}\:{f}\left({x}\right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{72}{x}−\mathrm{8}}+\sqrt{\mathrm{32}{x}−\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{128}{x}+\mathrm{240}}=\mathrm{5} \\ $$$${for}\:{f}\left({x}\right)\in\mathbb{R}\Rightarrow \\ $$$$\mathrm{32}{x}−\mathrm{4}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{128}{x}+\mathrm{240}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\frac{\mathrm{240}}{\mathrm{128}} \\ $$$$\Rightarrow{x}\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${try}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{72}{x}−\mathrm{8}}+\sqrt{\mathrm{32}{x}−\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{128}{x}+\mathrm{240}}= \\ $$$$=\sqrt[{\mathrm{3}}]{\mathrm{9}−\mathrm{8}}+\sqrt{\mathrm{4}−\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{16}+\mathrm{240}}= \\ $$$$=\mathrm{1}+\mathrm{0}+\mathrm{4}=\mathrm{5} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{8}}\:{is}\:{a}\:{solution} \\ $$$$\Rightarrow{the}\:{only}\:{solution}\:{of}\:{f}\left({x}\right)=\frac{\mathrm{5}}{\mathrm{2}}\:{over}\:\mathbb{R} \\ $$$${is}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by HongKing last updated on 25/Dec/21

cool my dear Sir thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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