Answer:
[tex]x = (\frac{-3}{4}+\frac{-1}{4}\sqrt{5} ) \:\:and\:\: (3 + \sqrt{5})[/tex]
Step-by-step explanation:
For a quadratic equation ax² + bx + c ,
its roots are = [tex]x = \frac{-b + \sqrt{b^2 - 4ac} }{2a} \:\: and \:\: \frac{-b - \sqrt{b^2 - 4ac} }{2a}[/tex]
In the question ,
Quadratic equation = (√5 - 3)x² + 3x + (√5 + 3)
By using quadratic formula ,
[tex]x = \frac{-3 + \sqrt{3^2 - 4\times(\sqrt{5}-3 )\times(\sqrt{5}+3 ) } }{2(\sqrt{5} -3)} ; \frac{-3 - \sqrt{3^2 - 4 \times (\sqrt{5}-3 )(\sqrt{5} +3)} }{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{-3 + \sqrt{9 - 4\times(\sqrt{5}^2-3^2 )} }{2(\sqrt{5} -3)} ; \frac{-3 - \sqrt{9 - 4 \times (\sqrt{5}^2-3^2 )}}{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{-3 + \sqrt{9 - 4\times(-4)} }{2(\sqrt{5} -3)} ; \frac{-3 - \sqrt{9 - 4 \times (-4)}}{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{-3 + \sqrt{9 +16} }{2(\sqrt{5} -3)} ; \frac{-3 - \sqrt{9 +16}}{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{-3 + 5 }{2(\sqrt{5} -3)} ; \frac{-3 -5}{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{2 }{2(\sqrt{5} -3)} ; \frac{-8}{2(\sqrt{5} -3)}[/tex]
[tex]= \frac{1 }{(\sqrt{5} -3)} ; \frac{-4}{(\sqrt{5} -3)}[/tex]
Rationalizing the roots ,
[tex]x = \frac{1 \times (\sqrt{5} + 3) }{(\sqrt{5} -3) \times (\sqrt{5} + 3)} ; \frac{-4 \times (\sqrt{5} +3)}{(\sqrt{5} -3)(\sqrt{5} +3)}[/tex]
[tex]= \frac{\sqrt{5} + 3}{-4} ; \frac{-4(\sqrt{5} +3)}{-4}[/tex]
[tex]= (\frac{-3}{4}+\frac{-1}{4}\sqrt{5} ) ; (3 + \sqrt{5})[/tex]
