solve-for-x-2-x-2-x-2-x-2-x-3-

Question Number 62489 by Tawa1 last updated on 21/Jun/19

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:+\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}{\:\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:−\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}\:\:=\:\:\mathrm{3} \\ $$

Answered by som(math1967) last updated on 22/Jun/19

(((√(2−x))+(√(2+x))+(√(2−x))−(√(2+x)))/( (√(2−x))+(√(2+x))−(√(2−x))+(√(2+x))))=((3+1)/(3−1)) ★ ((2(√(2−x)))/(2(√(2+x))))=(4/2) ((2−x)/(2+x))=4 ★★ 8+4x=2−x⇒x=−(6/5)ans ★using componendo and dividendo ★★squaring both side

$$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}+\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}−\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:\:\bigstar \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}−{x}}}{\mathrm{2}\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4}\:\:\bigstar\bigstar \\ $$$$\mathrm{8}+\mathrm{4}{x}=\mathrm{2}−{x}\Rightarrow{x}=−\frac{\mathrm{6}}{\mathrm{5}}{ans} \\ $$$$\bigstar{using}\:{componendo}\:{and}\:{dividendo} \\ $$$$\bigstar\bigstar{squaring}\:{both}\:{side} \\ $$

Commented by MJS last updated on 22/Jun/19

you should make your first step clearer. in fact there are a few steps: (((√(2−x))+(√(2+x)))/( (√(2−x))−(√(2+x))))=3 (√(2−x))+(√(2+x))=3(√(2−x))−3(√(2+x)) 4(√(2+x))=2(√(2−x)) 2(√(2+x))=(√(2−x)) 4(2+x)=2−x 6+5x=0 x=−(6/5)

$$\mathrm{you}\:\mathrm{should}\:\mathrm{make}\:\mathrm{your}\:\mathrm{first}\:\mathrm{step}\:\mathrm{clearer}. \\ $$$$\mathrm{in}\:\mathrm{fact}\:\mathrm{there}\:\mathrm{are}\:\mathrm{a}\:\mathrm{few}\:\mathrm{steps}: \\ $$$$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}=\mathrm{3}\sqrt{\mathrm{2}−{x}}−\mathrm{3}\sqrt{\mathrm{2}+{x}} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}+{x}}=\mathrm{2}\sqrt{\mathrm{2}−{x}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}+{x}}=\sqrt{\mathrm{2}−{x}} \\ $$$$\mathrm{4}\left(\mathrm{2}+{x}\right)=\mathrm{2}−{x} \\ $$$$\mathrm{6}+\mathrm{5}{x}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Commented by peter frank last updated on 22/Jun/19

$${nice}\:{too} \\ $$

Answered by MJS last updated on 21/Jun/19

defined for 2−x≥0∧2+x≥0∧(√(2−x))≠(√(2+x)) ⇒ −2≤x≤2∧x≠0 ((((√(2−x))+(√(2+x)))^2 )/(((√(2−x))−(√(2+x)))((√(2−x))+(√(2+x)))))=3 −((2+(√(2−x))(√(2+x)))/x)=3 (√(2−x))(√(2+x))=−3x−2 4−x^2 =9x^2 +12x+4 10x^2 +12x=0 x(5x+6)=0 x_1 =0 [not valid] x_2 =−(6/5)

$$\mathrm{defined}\:\mathrm{for} \\ $$$$\mathrm{2}−{x}\geqslant\mathrm{0}\wedge\mathrm{2}+{x}\geqslant\mathrm{0}\wedge\sqrt{\mathrm{2}−{x}}\neq\sqrt{\mathrm{2}+{x}} \\ $$$$\Rightarrow\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2}\wedge{x}\neq\mathrm{0} \\ $$$$\frac{\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}\right)\left(\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}\right)}=\mathrm{3} \\ $$$$−\frac{\mathrm{2}+\sqrt{\mathrm{2}−{x}}\sqrt{\mathrm{2}+{x}}}{{x}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{2}−{x}}\sqrt{\mathrm{2}+{x}}=−\mathrm{3}{x}−\mathrm{2} \\ $$$$\mathrm{4}−{x}^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{4} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} +\mathrm{12}{x}=\mathrm{0} \\ $$$${x}\left(\mathrm{5}{x}+\mathrm{6}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:\left[\mathrm{not}\:\mathrm{valid}\right] \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Commented by Tawa1 last updated on 21/Jun/19

$$\mathrm{Good}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 22/Jun/19

the path of som (math67) is better, I didn′t realize this possibility at first (always looking for trouble, my thinking is too complicated sometimes)

$$\mathrm{the}\:\mathrm{path}\:\mathrm{of}\:{som}\:\left({math}\mathrm{67}\right)\:\mathrm{is}\:\mathrm{better},\:\mathrm{I}\:\mathrm{didn}'\mathrm{t} \\ $$$$\mathrm{realize}\:\mathrm{this}\:\mathrm{possibility}\:\mathrm{at}\:\mathrm{first}\:\left(\mathrm{always}\:\mathrm{looking}\right. \\ $$$$\mathrm{for}\:\mathrm{trouble},\:\mathrm{my}\:\mathrm{thinking}\:\mathrm{is}\:\mathrm{too}\:\mathrm{complicated} \\ $$$$\left.\mathrm{sometimes}\right) \\ $$

Commented by Tawa1 last updated on 22/Jun/19

$$\mathrm{Hahahaha} \\ $$

Commented by peter frank last updated on 22/Jun/19

$${thank}\:{you}\: \\ $$

Answered by Rasheed.Sindhi last updated on 22/Jun/19

AnOtherWay (((√(2 − x)) + (√(2 + x)))/( (√(2 − x)) − (√(2 + x)))) = 3 ⇒((((√(2−x))/( (√(2+x))))+1)/(((√(2−x))/( (√(2+x))))−1))=3 ⇒((((√(2−x))/( (√(2+x))))+1)/(((√(2−x))/( (√(2+x))))−1))=(3/1)=((2+1)/(2−1)) ((√(2−x))/( (√(2+x))))=2 ((2−x)/(2+x))=4 8+4x=2−x 5x=−6 x=−6/5

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{W}\mathrm{ay} \\ $$$$\:\:\:\:\frac{\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:+\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}{\:\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:−\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}\:\:=\:\:\mathrm{3} \\ $$$$\:\:\:\:\Rightarrow\frac{\frac{\sqrt{\mathrm{2}−{x}}}{\:\sqrt{\mathrm{2}+{x}}}+\mathrm{1}}{\frac{\sqrt{\mathrm{2}−{x}}}{\:\sqrt{\mathrm{2}+{x}}}−\mathrm{1}}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\frac{\frac{\sqrt{\mathrm{2}−{x}}}{\:\sqrt{\mathrm{2}+{x}}}+\mathrm{1}}{\frac{\sqrt{\mathrm{2}−{x}}}{\:\sqrt{\mathrm{2}+{x}}}−\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{1}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}−{x}}}{\:\sqrt{\mathrm{2}+{x}}}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}+\mathrm{4}{x}=\mathrm{2}−{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{x}=−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=−\mathrm{6}/\mathrm{5} \\ $$

Commented by Tawa1 last updated on 22/Jun/19