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How do you find the quadratic equation if the roots are \(x = 3 + \sqrt{5}\) and\(x = 3 - \sqrt{5}\)?

 Apr 15, 2020
 #1
avatar+23689 
+2

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  

 Apr 15, 2020
 #2
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0

 

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

.

 Apr 15, 2020
 #3
avatar+109524 
+1

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  cheeky

 

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.     wink

 Apr 16, 2020
 #4
avatar+24995 
+1

How do you find the quadratic equation if the roots are \(x = 3 + \sqrt{5}\) and \(x = 3 - \sqrt{5}\)?

 

\(\begin{array}{|rcll|} \hline && (x-r_1)(x-r_2) \\ &=& \Big(x-(3 - \sqrt{5})\Big)\Big(x-(3 + \sqrt{5})\Big) \\ &=& (x-3 + \sqrt{5})(x-3 - \sqrt{5}) \\ &=& \Big( (x-3) + \sqrt{5}\Big) \Big((x-3) - \sqrt{5}\Big) \\ &=& (x-3)^2 - (\sqrt{5})^2 \\ &=& x^2-6x+9-5 \\ &=& \mathbf{ x^2-6x+4 } \\ \hline \end{array}\)

 

laugh

 Apr 16, 2020

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