122-a-2-123-a-2-

Question Number 179775 by Acem last updated on 02/Nov/22

$$\:\left(\mathrm{122}−{a}\right)^{\mathrm{2}} =\:\left(\mathrm{123}−{a}\right)^{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 02/Nov/22

(122−a)^2 = (123−a)^2 (122−a)^2 − (122−a+1)^2 =0 (122−a)^2 −(122−a)^2 −1−2(122−a)=0 −1−244+2a=0 a=((245)/2)

$$\:\left(\mathrm{122}−{a}\right)^{\mathrm{2}} =\:\left(\mathrm{123}−{a}\right)^{\mathrm{2}} \\ $$$$\:\left(\mathrm{122}−{a}\right)^{\mathrm{2}} −\:\left(\mathrm{122}−{a}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\cancel{\left(\mathrm{122}−{a}\right)^{\mathrm{2}} }−\cancel{\left(\mathrm{122}−{a}\right)^{\mathrm{2}} }−\mathrm{1}−\mathrm{2}\left(\mathrm{122}−{a}\right)=\mathrm{0} \\ $$$$−\mathrm{1}−\mathrm{244}+\mathrm{2}{a}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{245}}{\mathrm{2}} \\ $$

Commented by Acem last updated on 02/Nov/22

Thanks Sir! Correct but a bit complex a^2 −b^2 = (a−b)(a+b) was easier

$${Thanks}\:{Sir}! \\ $$$${Correct}\:{but}\:{a}\:{bit}\:{complex} \\ $$$$\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\:\left({a}−{b}\right)\left({a}+{b}\right)\:{was}\:{easier} \\ $$

Answered by Rasheed.Sindhi last updated on 02/Nov/22

(122−a)^2 = (123−a)^2 { ((122−a=123−a⇒Absurd ⌢^(• •) )),((or)),((122−a=a−123=2a=122+123=245✓⌣^(• •) )) :} a=((245)/2)

$$\:\left(\mathrm{122}−{a}\right)^{\mathrm{2}} =\:\left(\mathrm{123}−{a}\right)^{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{122}−{a}=\mathrm{123}−{a}\Rightarrow{Absurd}\:\:\:\overset{\bullet\:\:\:\:\:\bullet} {\frown}}\\{\mathrm{or}}\\{\mathrm{122}−{a}={a}−\mathrm{123}=\mathrm{2}{a}=\mathrm{122}+\mathrm{123}=\mathrm{245}\checkmark\overset{\bullet\:\:\:\:\:\bullet} {\smile}}\end{cases} \\ $$$${a}=\frac{\mathrm{245}}{\mathrm{2}} \\ $$

Commented by Acem last updated on 02/Nov/22

I hope you always being^• ⌣^• i coudn′t make like yours :/

$${I}\:{hope}\:{you}\:{always}\:{being}\:^{\bullet} \smile^{\bullet} \: \\ $$$$\:{i}\:{coudn}'{t}\:{make}\:{like}\:{yours}\::/ \\ $$

Answered by Frix last updated on 02/Nov/22

(x−1)^2 =x^2 x=(1/2) 123−a=(1/2) a=((245)/2)

$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{123}−{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{245}}{\mathrm{2}} \\ $$

Commented by Acem last updated on 02/Nov/22

Hi friend, you made root for each side and took what you like :) “−x” what about +x? Anyway check out our friends′ methods Thank you!

$${Hi}\:{friend},\:{you}\:{made}\:{root}\:{for}\:{each}\:{side}\:{and}\:{took} \\ $$$$\left.\:{what}\:{you}\:{like}\::\right)\:\:“−{x}''\:{what}\:{about}\:+{x}? \\ $$$${Anyway}\:{check}\:{out}\:{our}\:{friends}'\:{methods} \\ $$$${Thank}\:{you}! \\ $$

Commented by Rasheed.Sindhi last updated on 03/Nov/22

I think FrΠ^(⧫) x sir has not made root of both sides: (x−1)^2 =x^2 x^2 −2x+1=x^2 −2x+1=0 x=(1/2)

$${I}\:{think}\:{Fr}\overset{\blacklozenge} {\Pi}{x}\:{sir}\:{has}\:{not}\:{made}\:{root}\:{of} \\ $$$${both}\:{sides}: \\ $$$$\:\:\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}={x}^{\mathrm{2}} \\ $$$$\:\:\:\:−\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 03/Nov/22

(((123−a)/(122−a)))^2 =1 (((123−a)/(122−a))−1+1)^2 =1 ((1/(122−a))+1)^2 =1 ((1/(122−a)))^2 +1^(×) +(2/(122−a))=1^(×) (1/((122−a)^2 ))+(2/(122−a))=0 1+2(122−a)=0 [Multiply by (122−a)^2 ] a=((245)/2)

$$\left(\frac{\mathrm{123}−{a}}{\mathrm{122}−{a}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{123}−{a}}{\mathrm{122}−{a}}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{122}−{a}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{122}−{a}}\right)^{\mathrm{2}} +\overset{×} {\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{122}−{a}}=\overset{×} {\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{122}−{a}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{122}−{a}}=\mathrm{0} \\ $$$$\:\:\mathrm{1}+\mathrm{2}\left(\mathrm{122}−{a}\right)=\mathrm{0}\:\:\left[{Multiply}\:{by}\:\left(\mathrm{122}−{a}\right)^{\mathrm{2}} \right] \\ $$$${a}=\frac{\mathrm{245}}{\mathrm{2}} \\ $$

Commented by Acem last updated on 02/Nov/22

A strange way, but it seems interesting hmmm what happend from ((1/(122−a)))^2 +1^(×) +(2/(122−a))=1^(×) to this: 1+2(122−a)=0

$${A}\:{strange}\:{way},\:{but}\:{it}\:{seems}\:{interesting} \\ $$$$ \\ $$$${hmmm}\:{what}\:{happend}\:{from} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{122}−{a}}\right)^{\mathrm{2}} +\overset{×} {\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{122}−{a}}=\overset{×} {\mathrm{1}} \\ $$$$\:{to}\:{this}: \\ $$$$\:\:\mathrm{1}+\mathrm{2}\left(\mathrm{122}−{a}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$

Commented by Acem last updated on 02/Nov/22

$${i}\:{got}\:{it}\:….\:{thanksss} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Nov/22

•Thank you for commenting in such a friendly way! •You got it but now I insert one step more for all others.

$$\bullet\mathcal{T}{hank}\:{you}\:{for}\:{commenting}\:{in}\:{such} \\ $$$${a}\:{friendly}\:{way}! \\ $$$$\bullet{You}\:{got}\:{it}\:{but}\:{now}\:{I}\:{insert}\:{one}\:{step} \\ $$$$\:\:\:{more}\:{for}\:{all}\:{others}. \\ $$

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