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Find positive integers \((a,b)\) so that \(\sqrt{37 + 20 \sqrt{3}} = a + b \sqrt{3}.\)

 Apr 15, 2019

Best Answer 

 #1
avatar+23786 
+2

Find positive integers \((a,b)\) so that \(\sqrt{37 + 20 \sqrt{3}} = a + b \sqrt{3}\).

 

\(\begin{array}{|rcll|} \hline \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{a + b \sqrt{3}} \\\\ 37 + 20 \sqrt{3} &=& \left(a + b \sqrt{3}\right)^2 \\ 37 + 20 \sqrt{3} &=& a^2+2ab\sqrt{3} +3b^2 \\ 37 + 20 \sqrt{3} &=& \underbrace{a^2+3b^2}_{=37}+\underbrace{2ab}_{=20} \sqrt{3} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & 2ab &=& 20 \\ & ab &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{10}{a}} \\ \hline (2) & a^2+3b^2 &=& 37 \\ & a^2+3\dfrac{10^2}{a^2} &=& 37 \\ & a^2+ \dfrac{300}{a^2} &=& 37 \quad |\quad \cdot a^2 \\ & a^4+ 300 &=& 37a^2 \\ & a^4-37a^2+ 300 &=& 0 \\\\ & a^2 &=& \dfrac{37 \pm \sqrt{37^2-4\cdot 300} }{2} \\ & a^2 &=& \dfrac{37 \pm \sqrt{169} }{2} \\ & a^2 &=& \dfrac{37 \pm 13 }{2} \\\\ & a^2 = 25 &\text{or}& a^2 = 12 \\ & a = \pm 5 &\text{or}& a = \pm 2\sqrt{3} \quad |\quad \text{positive integers } (a,b) \\ & a = 5 &\text{or}& a = 2\sqrt{3} \quad |\quad a \text{ without } \sqrt{3} \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b &=& \dfrac{10}{a} \\ & b &=& \dfrac{10}{5} \\ & \mathbf{b} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{5 + 2 \sqrt{3}} \\ \end{array}\)

 

laugh

 Apr 15, 2019
 #1
avatar+23786 
+2
Best Answer

Find positive integers \((a,b)\) so that \(\sqrt{37 + 20 \sqrt{3}} = a + b \sqrt{3}\).

 

\(\begin{array}{|rcll|} \hline \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{a + b \sqrt{3}} \\\\ 37 + 20 \sqrt{3} &=& \left(a + b \sqrt{3}\right)^2 \\ 37 + 20 \sqrt{3} &=& a^2+2ab\sqrt{3} +3b^2 \\ 37 + 20 \sqrt{3} &=& \underbrace{a^2+3b^2}_{=37}+\underbrace{2ab}_{=20} \sqrt{3} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & 2ab &=& 20 \\ & ab &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{10}{a}} \\ \hline (2) & a^2+3b^2 &=& 37 \\ & a^2+3\dfrac{10^2}{a^2} &=& 37 \\ & a^2+ \dfrac{300}{a^2} &=& 37 \quad |\quad \cdot a^2 \\ & a^4+ 300 &=& 37a^2 \\ & a^4-37a^2+ 300 &=& 0 \\\\ & a^2 &=& \dfrac{37 \pm \sqrt{37^2-4\cdot 300} }{2} \\ & a^2 &=& \dfrac{37 \pm \sqrt{169} }{2} \\ & a^2 &=& \dfrac{37 \pm 13 }{2} \\\\ & a^2 = 25 &\text{or}& a^2 = 12 \\ & a = \pm 5 &\text{or}& a = \pm 2\sqrt{3} \quad |\quad \text{positive integers } (a,b) \\ & a = 5 &\text{or}& a = 2\sqrt{3} \quad |\quad a \text{ without } \sqrt{3} \\ & \mathbf{a} &\mathbf{=}& \mathbf{5} \\\\ & b &=& \dfrac{10}{a} \\ & b &=& \dfrac{10}{5} \\ & \mathbf{b} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} \mathbf{\sqrt{37 + 20 \sqrt{3}}} & \mathbf{=} & \mathbf{5 + 2 \sqrt{3}} \\ \end{array}\)

 

laugh

heureka Apr 15, 2019
 #2
avatar+106514 
+1

Excellent, heureka!!!!

 

cool cool cool

CPhill  Apr 15, 2019
 #3
avatar+23786 
+2

Thank you, CPhill !

 

laugh

heureka  Apr 15, 2019

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