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Question Number 215874 by MathematicalUser2357 last updated on 20/Jan/25

$$\mathrm{If}{x}^2+3x+2=y^2+5y+8,$$$$\mathrm{Prove}\mathrm{that}x=\frac{-3\pm \sqrt{4y^2+20y+33}}{2}.$$

Answered by MrGaster last updated on 20/Jan/25

$$x^2+3x+2=y^2+5y+8$$$$x^2+3x-y^2-5y-6=0$$$$=y^2+5x+{\left(\frac{5}{2}\right)}^2+6+{\left(\frac{3}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2$$$${(x+\frac{3}{2})}^2={(y+\frac{5}{2})}^2+\frac{9}{4}-\frac{25}{4}+6$$$${(x+\frac{3}{2})}^2={(y+\frac{5}{2})}^2+\frac{24-16}{4}$$$${(x+\frac{3}{2})}^2={(y+\frac{5}{2})}^2+\frac{8}{4}$$$${(x+\frac{3}{2})}^2={(y+\frac{5}{2})}^2+2$$$$x+\frac{3}{2}=\pm \sqrt{{(y+\frac{5}{2})}^2+2}$$$$x=\frac{-3\pm \sqrt{{(2y+5)}^2+8}}{2}$$$$x=\frac{-3\pm \sqrt{4y^2+20y+25+8}}{2}$$$$x=\frac{-3\pm \sqrt{4y^2+20y+33}}{2}$$$$[\mathrm{Q}.\mathrm{E}.\mathrm{D}]$$

Answered by Rasheed.Sindhi last updated on 20/Jan/25

$$x^2+3x+2=y^2+5y+8,$$$$x^2+3x-y^2-5y-6=0$$$$x=\frac{-\left(3\right)\pm \sqrt{{\left(3\right)}^2-4\left(1\right)(-y^2-5y-6)}}{2\left(1\right)}$$$$x=\frac{-3\pm \sqrt{9+4y^2+20y+24}}{2}$$$$x=\frac{-3\pm \sqrt{4y^2+20y+33}}{2}$$

Answered by MATHEMATICSAM last updated on 20/Jan/25

$$x^2+3x+2=y^2+5y+8$$$$\Rightarrow x^2+3x+2-y^2-5y-8=0$$$$\Rightarrow x^2+3x-(y^2+5y+6)=0$$$$x=\frac{-3\pm \sqrt{9+4y^2+20y+24}}{2}$$$$\Rightarrow x=\frac{-3\pm \sqrt{4y^2+20y+33}}{2}\left(\mathrm{Proved}\right)$$

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