How do you find the quadratic equation if the roots are \(x = 3 + \sqrt{5}\) and\(x = 3 - \sqrt{5}\)?
+36916 The equation would look like this
(x - r1)(x-r2) = x^2 - r1x - r2x + r1r2
= x^2 -3x-sqrt5x - 3x+sqrt5x + 9-5
= x^2 -6x + 4
Recognize that the two roots are in the form x+y and x–y. When this is the case, their product is always x2 – y2.
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+118608 How do you find the quadratic equation if the roots are
\(x = 3 + \sqrt{5} \;\;and\;\; x = 3 - \sqrt{5}\)
Thanks EP
I want to experiment with a different method 
The roots of a quadratic equation are
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ so\\ {-b \pm \sqrt{b^2-4ac} \over 2a} = 3 \pm \sqrt{5}\\ \frac{-b}{2a}=3\qquad and \qquad \frac{\sqrt{b^2-4ac}}{2a} =\sqrt5\\ b=-6a\qquad and \qquad \sqrt{b^2-4ac} =2a\sqrt5\\ b=-6a\qquad and \qquad b^2-4ac =4a^2*5\\ \boxed{b=-6a}\qquad and \qquad b^2-4ac =20a^2\\ (-6a)^2-4ac =20a^2\\ 36a^2-4ac =20a^2\\ 4a=c\\ Let\;\; a=1, \;\;then\;\;c=4\;\;and \;\;b=-6\\ \text{So the quadratic equation is}\\ y= x^2-6x+4 \)
EP's method was a lot easier. 
+26367 How do you find the quadratic equation if the roots are \(x = 3 + \sqrt{5}\) and \(x = 3 - \sqrt{5}\)?
\(\begin{array}{|rcll|} \hline && (x-r_1)(x-r_2) \\ &=& \Big(x-(3 - \sqrt{5})\Big)\Big(x-(3 + \sqrt{5})\Big) \\ &=& (x-3 + \sqrt{5})(x-3 - \sqrt{5}) \\ &=& \Big( (x-3) + \sqrt{5}\Big) \Big((x-3) - \sqrt{5}\Big) \\ &=& (x-3)^2 - (\sqrt{5})^2 \\ &=& x^2-6x+9-5 \\ &=& \mathbf{ x^2-6x+4 } \\ \hline \end{array}\)
