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Question Number 33764 by .none. last updated on 23/Apr/18

2x−y=(1/((√5)−2)) x+2y=(1/((√5)+2)) [3x^2 −8xy−3y^2 ]?

$$\mathrm{2}{x}−{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$${x}+\mathrm{2}{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}+\mathrm{2}} \\ $$$$\left[\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{xy}−\mathrm{3}{y}^{\mathrm{2}} \right]? \\ $$

Answered by Rasheed.Sindhi last updated on 23/Apr/18

3x^2 −8xy−3y^2 =(x−3y)(3x+y) −−−−−−−−− 2x−y=(1/((√5)−2))............A x+2y=(1/((√5)+2))..............B A+B: 3x+y=(1/((√5)−2))+(1/((√5)+2))=(((√5)+2+(√5)−2)/(5−4)) =2(√5) A−B: x−3y=(1/((√5)−2))−(1/((√5)+2))=(((√5)+2−(√5)+2)/(5−4)) =4 −−−−−−−−−−−− 3x^2 −8xy−3y^2 =(x−3y)(3x+y) =(4)(2(√5))=8(√5)

$$\mathrm{3x}^{\mathrm{2}} −\mathrm{8xy}−\mathrm{3y}^{\mathrm{2}} =\left(\mathrm{x}−\mathrm{3y}\right)\left(\mathrm{3x}+\mathrm{y}\right) \\ $$$$−−−−−−−−− \\ $$$$\mathrm{2}{x}−{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}−\mathrm{2}}............\mathrm{A} \\ $$$${x}+\mathrm{2}{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}+\mathrm{2}}..............\mathrm{B} \\ $$$$\mathrm{A}+\mathrm{B}:\:\mathrm{3x}+\mathrm{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}−\mathrm{2}}+\frac{\mathrm{1}}{\sqrt{\mathrm{5}}+\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{2}+\sqrt{\mathrm{5}}−\mathrm{2}}{\mathrm{5}−\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{A}−\mathrm{B}:\:\mathrm{x}−\mathrm{3y}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}−\mathrm{2}}−\frac{\mathrm{1}}{\sqrt{\mathrm{5}}+\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{2}−\sqrt{\mathrm{5}}+\mathrm{2}}{\mathrm{5}−\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4} \\ $$$$−−−−−−−−−−−− \\ $$$$\mathrm{3x}^{\mathrm{2}} −\mathrm{8xy}−\mathrm{3y}^{\mathrm{2}} =\left(\mathrm{x}−\mathrm{3y}\right)\left(\mathrm{3x}+\mathrm{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{4}\right)\left(\mathrm{2}\sqrt{\mathrm{5}}\right)=\mathrm{8}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$ \\ $$

Commented by .none. last updated on 23/Apr/18

Thanks to you,i can solve this now

$${Thanks}\:{to}\:{you},{i}\:{can}\:{solve}\:{this}\:{now} \\ $$

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