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Question Number 172031 by Mikenice last updated on 23/Jun/22

solve: (√(x^2 −4x+4)) +(√(x^2 +4x+4)) =(√(x^2 −6x+9))

$${solve}: \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}\:\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$

Answered by puissant last updated on 23/Jun/22

⇒(x−2)+(x+2)=(√(x^2 −6x+9)) ⇒ 4x^2 =x^2 −6x+9 ⇒ 3x^2 +6x−9=0 ⇒ x^2 +2x−3=0 Δ=4−4(1)(−3)=16 → (√Δ)=4. x_1 =−3 and x_2 = 1 take S={ −3 }

$$\Rightarrow\left({x}−\mathrm{2}\right)+\left({x}+\mathrm{2}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$$$\Rightarrow\:\mathrm{4}{x}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9} \\ $$$$\Rightarrow\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\Delta=\mathrm{4}−\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{3}\right)=\mathrm{16}\:\rightarrow\:\sqrt{\Delta}=\mathrm{4}. \\ $$$${x}_{\mathrm{1}} =−\mathrm{3}\:\:\:{and}\:\:\:{x}_{\mathrm{2}} =\:\mathrm{1} \\ $$$$\:\:\:\:\:\:{take}\:\:{S}=\left\{\:−\mathrm{3}\:\right\} \\ $$

Commented by Mikenice last updated on 23/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Jun/22

x=1 is not valid.

$${x}=\mathrm{1}\:{is}\:{not}\:{valid}. \\ $$

Answered by Rasheed.Sindhi last updated on 23/Jun/22

(√(x^2 −4x+4)) +(√(x^2 +4x+4)) =(√(x^2 −6x+9)) (x+2)+(x−2)=x−3 2x=x−3 x=−3

$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}\:\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}} \\ $$$$\left({x}+\mathrm{2}\right)+\left({x}−\mathrm{2}\right)={x}−\mathrm{3} \\ $$$$\mathrm{2}{x}={x}−\mathrm{3} \\ $$$${x}=−\mathrm{3} \\ $$

Answered by kapoorshah last updated on 23/Jun/22

(√((x − 2)^2 )) + (√((x + 2)^2 )) = (√((x − 3)^2 )) ∣x − 2∣ + ∣x + 2∣ = ∣x − 3∣ x < − 2 ⇒ − x + 2 − x − 2 = − x + 3 x = − 3 − 2 ≤ x < 2 ⇒ − x + 2 + x + 2 = − x + 3 x = − 1 x ≥ 3 ⇒ x − 2 + x + 2 = x − 3 x = − 3 (rejected) solution : − 3, 1

$$\sqrt{\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} \:}\:+\:\sqrt{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:}\:=\:\sqrt{\left({x}\:−\:\mathrm{3}\right)^{\mathrm{2}} }\: \\ $$$$\mid{x}\:−\:\mathrm{2}\mid\:+\:\mid{x}\:+\:\mathrm{2}\mid\:=\:\mid{x}\:−\:\mathrm{3}\mid \\ $$$$ \\ $$$${x}\:<\:−\:\mathrm{2}\: \\ $$$$\Rightarrow\:−\:{x}\:+\:\mathrm{2}\:−\:{x}\:−\:\mathrm{2}\:=\:−\:{x}\:+\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:{x}\:=\:−\:\mathrm{3} \\ $$$$ \\ $$$$−\:\mathrm{2}\:\leqslant\:{x}\:<\:\mathrm{2}\: \\ $$$$\Rightarrow\:−\:{x}\:+\:\mathrm{2}\:+\:{x}\:+\:\mathrm{2}\:=\:−\:{x}\:+\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:{x}\:=\:−\:\mathrm{1} \\ $$$$ \\ $$$${x}\:\geqslant\:\mathrm{3}\: \\ $$$$\Rightarrow\:{x}\:−\:\mathrm{2}\:+\:{x}\:+\:\mathrm{2}\:=\:{x}\:−\:\mathrm{3} \\ $$$$\:\:\:\:\:\:{x}\:=\:−\:\mathrm{3}\:\left({rejected}\right) \\ $$$$ \\ $$$${solution}\::\:−\:\mathrm{3},\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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