We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
62
3
avatar

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

 Apr 15, 2019

Best Answer 

 #1
avatar+22290 
+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+22290 
+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+101161 
+1

Excellent, heureka!!!!

 

cool cool cool

CPhill  Apr 15, 2019
 #3
avatar+22290 
+2

Thank you, CPhill !

 

laugh

heureka  Apr 15, 2019

13 Online Users

avatar