Question Number 138193 by mey3nipaba last updated on 10/Apr/21
$${Given}\:{x}\neq{y}\:{and}\:{x}^{\mathrm{2}} =\mathrm{25}{x}+{y},\:{y}^{\mathrm{2}} ={x}+\mathrm{25}{y}\: \\ $$$${solve}\:{for}\:{the}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\:{without}\: \\ $$$${using}\:{calculators}\:{or}\:{tools}. \\ $$$${Show}\:{your}\:{method}. \\ $$
Answered by liberty last updated on 11/Apr/21
![x^2 =25x+y y^2 =x+25y ⇒x^2 +y^2 =26(x+y) ⇒x^2 −y^2 =24(x−y) ⇒(x−y)[(x+y)−24]=0 ⇒x+y = 24 (√(x^2 +y^2 +1)) = (√(26.24+1)) =(√((25+1)(25−1)+1)) =(√(25^2 )) = 25](https://www.tinkutara.com/question/Q138195.png)
$${x}^{\mathrm{2}} =\mathrm{25}{x}+{y} \\ $$$${y}^{\mathrm{2}} ={x}+\mathrm{25}{y} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{26}\left({x}+{y}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{24}\left({x}−{y}\right) \\ $$$$\Rightarrow\left({x}−{y}\right)\left[\left({x}+{y}\right)−\mathrm{24}\right]=\mathrm{0} \\ $$$$\Rightarrow{x}+{y}\:=\:\mathrm{24}\: \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\:=\:\sqrt{\mathrm{26}.\mathrm{24}+\mathrm{1}} \\ $$$$=\sqrt{\left(\mathrm{25}+\mathrm{1}\right)\left(\mathrm{25}−\mathrm{1}\right)+\mathrm{1}} \\ $$$$=\sqrt{\mathrm{25}^{\mathrm{2}} }\:=\:\mathrm{25} \\ $$
Commented by mey3nipaba last updated on 11/Apr/21
$${thank}\:{you}\:{very}\:{much}.\:{I}\:{appreciate}. \\ $$
Commented by mey3nipaba last updated on 11/Apr/21
$${But}\:{please}\:{I}\:{do}\:{not}\:{understand}\:{from}\:{where} \\ $$$${you}\:{equated}\:{it}\:{to}\:\mathrm{0}.\:{Can}\:{you}\:{explain}? \\ $$
Commented by liberty last updated on 11/Apr/21
$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −\mathrm{24}\left({x}−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}+{y}\right)−\mathrm{24}\left({x}−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}+{y}−\mathrm{24}\right)=\mathrm{0} \\ $$$${since}\:{x}\neq{y}\:{it}\:{follows}\:{that}\:{x}+{y}−\mathrm{24}=\mathrm{0} \\ $$
Commented by mey3nipaba last updated on 11/Apr/21
$${Thanks}.\:{I}\:{get}\:{this}\:{now}\:{but}\:{the}\:{part}\:{where} \\ $$$${you}\:{had}\:\mathrm{26}.\mathrm{4}\:{as}\:{the}\:{value}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} . \\ $$$${Isn}'{t}\:{it}\:{supposed}\:{to}\:{be}\:\mathrm{264}? \\ $$