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Question Number 168335 by vbr last updated on 08/Apr/22

(x/2)+4(√(x ))+6+3+1=x

$$\frac{{x}}{\mathrm{2}}+\mathrm{4}\sqrt{{x}\:}+\mathrm{6}+\mathrm{3}+\mathrm{1}={x} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Apr/22

(x/2)+4(√(x ))+10=x  x+8(√x) +20=2x  8(√x) +20=x  8(√x) =x−20  64x=x^2 −40x+400  x^2 −104x+400=0  (x−100)(x−4)=0  x=100 ∣ x=4  x=100 is a valid root.

$$\frac{{x}}{\mathrm{2}}+\mathrm{4}\sqrt{{x}\:}+\mathrm{10}={x} \\ $$$${x}+\mathrm{8}\sqrt{{x}}\:+\mathrm{20}=\mathrm{2}{x} \\ $$$$\mathrm{8}\sqrt{{x}}\:+\mathrm{20}={x} \\ $$$$\mathrm{8}\sqrt{{x}}\:={x}−\mathrm{20} \\ $$$$\mathrm{64}{x}={x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{400} \\ $$$${x}^{\mathrm{2}} −\mathrm{104}{x}+\mathrm{400}=\mathrm{0} \\ $$$$\left({x}−\mathrm{100}\right)\left({x}−\mathrm{4}\right)=\mathrm{0} \\ $$$${x}=\mathrm{100}\:\mid\:{x}=\mathrm{4} \\ $$$${x}=\mathrm{100}\:{is}\:{a}\:{valid}\:{root}. \\ $$

Answered by mr W last updated on 08/Apr/22

x−8(√x)−20=0  ((√x)+2)((√x)−10)=0  ⇒(√x)−10=0  ⇒x=100

$${x}−\mathrm{8}\sqrt{{x}}−\mathrm{20}=\mathrm{0} \\ $$$$\left(\sqrt{{x}}+\mathrm{2}\right)\left(\sqrt{{x}}−\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{x}}−\mathrm{10}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{100} \\ $$

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