Question Number 223161 by Rojarani last updated on 16/Jul/25
$${Solve}\:{for}\:{x}\:{if} \\ $$$$\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=\mathrm{1} \\ $$
Commented by fantastic last updated on 16/Jul/25
$${Why}\:{am}\:{i}\:{getting}\:\frac{\mathrm{5}}{\mathrm{4}}?? \\ $$
Commented by Rojarani last updated on 16/Jul/25
$$\:{Sir},\:{x}=\frac{\mathrm{5}}{\mathrm{4}}\:{not}\:{satisfying}. \\ $$
Commented by fantastic last updated on 16/Jul/25
$${yes} \\ $$
Commented by mr W last updated on 16/Jul/25
$${for}\:{x}\in{R}: \\ $$$${x}\geqslant\mathrm{1} \\ $$$${both}\:\sqrt{{x}+\mathrm{1}}\:{and}\:\sqrt{{x}−\mathrm{1}}\:{are}\:{strictly} \\ $$$${increasing},\:{that}\:{means} \\ $$$${f}\left({x}\right)=\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}\:{has}\:{mininum} \\ $$$${at}\:{x}=\mathrm{1},\:{f}\left({x}\right)_{{min}} ={f}\left(\mathrm{1}\right)=\sqrt{\mathrm{2}}>\mathrm{1} \\ $$$${therefore}\:{f}\left({x}\right)=\mathrm{1}\:{has}\:{no}\:{real}\:{root}. \\ $$
Answered by fantastic last updated on 16/Jul/25
![I think no solution exists here When x∈R (√(x−1))≥0 or x−1≥0 or x≥1 But when x=1 then (√(1+1))+(√(1−1))=(√2)≠1 So x>1 So (√(x+1))>0 and (√(x−1))>0...(i) Now (√(x+1))+(√(x−1))=1 So (√(x+1))<1[(i)] so x+1<1 or x<0 But x>1 So no solution exists](https://www.tinkutara.com/question/Q223168.png)
$${I}\:{think}\:{no}\:{solution}\:{exists}\:{here} \\ $$$${When}\:{x}\in\mathbb{R} \\ $$$$\sqrt{{x}−\mathrm{1}}\geqslant\mathrm{0} \\ $$$${or}\:{x}−\mathrm{1}\geqslant\mathrm{0} \\ $$$${or}\:{x}\geqslant\mathrm{1} \\ $$$${But}\:{when}\:{x}=\mathrm{1}\:{then}\: \\ $$$$\sqrt{\mathrm{1}+\mathrm{1}}+\sqrt{\mathrm{1}−\mathrm{1}}=\sqrt{\mathrm{2}}\neq\mathrm{1} \\ $$$${So}\:{x}>\mathrm{1} \\ $$$${So}\:\sqrt{{x}+\mathrm{1}}>\mathrm{0}\:{and}\:\sqrt{{x}−\mathrm{1}}>\mathrm{0}…\left(\mathrm{i}\right) \\ $$$${Now}\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=\mathrm{1} \\ $$$${So}\:\sqrt{{x}+\mathrm{1}}<\mathrm{1}\left[\left(\mathrm{i}\right)\right] \\ $$$${so}\:{x}+\mathrm{1}<\mathrm{1} \\ $$$${or}\:{x}<\mathrm{0} \\ $$$${But}\:{x}>\mathrm{1} \\ $$$${So}\:{no}\:{solution}\:{exists} \\ $$$$ \\ $$
Commented by Rojarani last updated on 16/Jul/25
$${Sir},\:{is}\:{there}\:{any}\:{imaginary}\:{roots}? \\ $$
Commented by fantastic last updated on 16/Jul/25
$${I}\:{dont}\:{have}\:{much}\:{knowledge}\:{about} \\ $$$${complex}\:{numbers}.{I}\:{cannot}\:{tell} \\ $$$${or}\:{assure}\:{you}\:{about}\:{that}. \\ $$$${But}\:{I}\:{think}\:{there}\:{is}\:{no}\:{imaginary}\:{roots} \\ $$
Commented by fantastic last updated on 16/Jul/25
Answered by Frix last updated on 16/Jul/25
$$\mathrm{No}\:\mathrm{complex}\:\mathrm{solution}: \\ $$$$\mathrm{Let}\:\sqrt{{x}+\mathrm{1}}={a}+{b}\mathrm{i}\wedge\sqrt{{x}−\mathrm{1}}={c}+{d}\mathrm{i}; \\ $$$$\:\:\:\:\:{a},\:{b},\:{c},\:{d}\:\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$$\left({a}+{c}\right)+\left({b}+{d}\right)\mathrm{i}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$${a}+{c}=\mathrm{1}\wedge{b}+{d}=\mathrm{0} \\ $$$${c}=\mathrm{1}−{a}\wedge{d}=−{b} \\ $$$$\sqrt{{x}+\mathrm{1}}={a}+{b}\mathrm{i} \\ $$$$\sqrt{{x}−\mathrm{1}}=\left(\mathrm{1}−{a}\right)−{b}\mathrm{i} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:{x}+\mathrm{1}=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+\mathrm{2}{ab}\mathrm{i} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{x}−\mathrm{1}=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}\right)+\mathrm{2}\left({a}−\mathrm{1}\right){b}\mathrm{i} \\ $$$$\mathrm{subtracting}\:\mathrm{both} \\ $$$$\mathrm{2}=\mathrm{2}{a}−\mathrm{1}+\mathrm{2}{b}\mathrm{i} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{a}−\mathrm{1}=\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{b}=\mathrm{0}\:\Rightarrow\:{b}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${c}=−\frac{\mathrm{1}}{\mathrm{2}}\wedge{d}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\sqrt{{x}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sqrt{{x}−\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{which}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\in\mathbb{C} \\ $$$$\mathrm{also}:\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:\mathrm{for}\:{x}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{equations}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right): \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{but} \\ $$$$\sqrt{{x}+\mathrm{1}}=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{correct} \\ $$$$\sqrt{{x}−\mathrm{1}}=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}\neq−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 16/Jul/25
![...which btw again shows why for r>0 ⇒ (√r)>0 [(√4)=2∧(√4)≠−2] Otherwise we could solve all of these: (√(x+1))+(√(x−1))=−2 (√(x+1))+(√(x−1))=−1 (√(x+1))+(√(x−1))=1 (√(x+1))+(√(x−1))=2 with x=(5/4) which would contradict everything we have learned since ≈800 years.](https://www.tinkutara.com/question/Q223177.png)
$$…\mathrm{which}\:\mathrm{btw}\:\mathrm{again}\:\mathrm{shows}\:\mathrm{why}\:\mathrm{for}\:{r}>\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{{r}}>\mathrm{0}\:\:\:\:\:\left[\sqrt{\mathrm{4}}=\mathrm{2}\wedge\sqrt{\mathrm{4}}\neq−\mathrm{2}\right] \\ $$$$\mathrm{Otherwise}\:\mathrm{we}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{all}\:\mathrm{of}\:\mathrm{these}: \\ $$$$\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=−\mathrm{2} \\ $$$$\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=−\mathrm{1} \\ $$$$\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=\mathrm{1} \\ $$$$\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}=\mathrm{2} \\ $$$$\mathrm{with}\:{x}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{which}\:\mathrm{would}\:\mathrm{contradict}\:\mathrm{everything}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{learned}\:\mathrm{since}\:\approx\mathrm{800}\:\mathrm{years}. \\ $$