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The solutions to the equation\(6x^2 + 10x = 4 - 10x - 6x^2 \ can\ be\ written\ in\ the form\\ x=\frac{P\pm \sqrt Q}{R}, where\ P\ and\ R\\ are\ relatively\ prime\ integers\\ and\ R>0. \ What \ is \ the\ product\ PQR ?\)

 

 Jun 17, 2019
 #1
avatar+23806 
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The solutions to the equation

\(6x^2 + 10x = 4 - 10x - 6x^2 \text{ can be written in the form }\\ x=\frac{P\pm \sqrt Q}{R}, \text{ where P and R}\\ \text{are relatively prime integers and } R>0. \\ \text{What is the product PQR} ? \)

 

\(\begin{array}{|rcll|} \hline 6x^2 + 10x &=& 4 - 10x - 6x^2 \\ 12x^2 +20x -4 &=& 0 \\\\ x &=& \dfrac{-20\pm\sqrt{20^2-4\cdot 12\cdot(-4) } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{400+4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4\cdot 100 +4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4}\sqrt{100+48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{148 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{4\cdot 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm2\sqrt{4}\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm 4\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-4\cdot 5\pm 4\sqrt{37 } } {4\cdot 6} \\ x &=& \dfrac{ 4\cdot (-5\pm \sqrt{37}) } {4\cdot 6} \\ \mathbf{x} &=& \mathbf{\dfrac{ -5\pm \sqrt{37} } {6} } \quad | \quad P=-5,\ Q=37,\ R = 6 \\\\ PQR&=& -5\cdot 37 \cdot 6 \\ \mathbf{PQR} &=& \mathbf{-1110 } \\ \hline \end{array}\)

 

Thank you Alan!

Thank you CPhill!

 

laugh

 Jun 17, 2019
edited by heureka  Jun 17, 2019
edited by heureka  Jun 18, 2019
edited by heureka  Jun 18, 2019
 #2
avatar+208 
+2

Thank You heureka!

NoobGuest  Jun 18, 2019

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