Question Number 215874 by MathematicalUser2357 last updated on 20/Jan/25
$$\mathrm{If}\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}={y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{8}, \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{33}}}{\mathrm{2}}. \\ $$
Answered by MrGaster last updated on 20/Jan/25
![x^2 +3x+2=y^2 +5y+8 x^2 +3x−y^2 −5y−6=0 =y^2 +5x+((5/2))^2 +6+((3/2))^2 −((5/2))^2 (x+(3/2))^2 =(y+(5/2))^2 +(9/4)−((25)/( 4))+6 (x+(3/2))^2 =(y+(5/2))^2 +((24−16)/4) (x+(3/2))^2 =(y+(5/2))^2 +(8/4) (x+(3/2))^2 =(y+(5/2))^2 +2 x+(3/2)=±(√((y+(5/2))^2 +2)) x=((−3±(√((2y+5)^2 +8)))/2) x=((−3±(√(4y^2 +20y+25+8)))/2) x=((−3±(√(4y^2 +20y+33)))/2) [Q.E.D]](https://www.tinkutara.com/question/Q215878.png)
$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}={y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{8} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}−{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{6}=\mathrm{0} \\ $$$$={y}^{\mathrm{2}} +\mathrm{5}{x}+\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{6}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({y}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{25}}{\:\mathrm{4}}+\mathrm{6} \\ $$$$\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({y}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{24}−\mathrm{16}}{\mathrm{4}} \\ $$$$\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({y}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{4}} \\ $$$$\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({y}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$${x}+\frac{\mathrm{3}}{\mathrm{2}}=\pm\sqrt{\left({y}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\left(\mathrm{2}{y}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{8}}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{25}+\mathrm{8}}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{33}}}{\mathrm{2}} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/25
$$\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}={y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{8}, \\ $$$$\:{x}^{\mathrm{2}} +\mathrm{3}{x}−{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{6}=\mathrm{0} \\ $$$${x}=\frac{−\left(\mathrm{3}\right)\pm\sqrt{\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{6}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{24}}}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{20}{y}+\mathrm{33}}}{\mathrm{2}} \\ $$
Answered by MATHEMATICSAM last updated on 20/Jan/25
$${x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{2}\:=\:{y}^{\mathrm{2}} \:+\:\mathrm{5}{y}\:+\:\mathrm{8} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{2}\:−\:{y}^{\mathrm{2}} \:−\:\mathrm{5}{y}\:−\:\mathrm{8}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:−\left({y}^{\mathrm{2}} \:+\:\mathrm{5}{y}\:+\:\mathrm{6}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{−\:\mathrm{3}\:\pm\:\sqrt{\mathrm{9}\:+\:\mathrm{4}{y}^{\mathrm{2}} \:+\:\mathrm{20}{y}\:+\:\mathrm{24}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\:=\:\frac{−\mathrm{3}\:\pm\:\sqrt{\mathrm{4}{y}^{\mathrm{2}} \:+\:\mathrm{20}{y}\:+\:\mathrm{33}}}{\mathrm{2}}\:\left(\mathrm{Proved}\right) \\ $$