+0  
 
0
34
1
avatar

There are two pairs (x,y) of real numbers that satisfy the equation x + y = 4xy = 5. Given that the solutions x are in the form x = (a \pm b sqrt(c))/d , where a, b, c, and d are positive integers and the expression is completely simplified, what is the value of a + b + c + d?

 May 30, 2021
 #1
avatar+25938 
+1

There are two pairs \((x,y)\) of real numbers that satisfy the equation \(x + y = 4xy = 5\).
Given that the solutions x are in the form \(x = \dfrac{a \pm b \sqrt{c}}{d}\) ,
where a, b, c, and d are positive integers and the expression is completely simplified,
what is the value of \(a + b + c + d\)?

 

My attempt:

\(\text{Let $a=x+y=5$ } \\ \text{Let $b=xy=\dfrac{5}{4}$ }\)

 

\(\begin{array}{|rcll|} \hline x^2 - Ax + B &=& 0 \\ x^2 - 5x + \dfrac{5}{4} &=& 0 \\\\ x &=& \dfrac{ 5 \pm \sqrt{25-4\dfrac{5}{4}} }{2} \\\\ x &=& \dfrac{5 \pm \sqrt{25-5} }{2} \\\ x &=& \dfrac{5 \pm \sqrt{20} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{5 \pm 2\sqrt{5} }{2}} \\ \hline x+y &=& 5 \\ y &=& 5-x \\ y &=& 5-\dfrac{5 \pm 2\sqrt{5} }{2} \\\ y &=& \dfrac{10-5 \mp 2\sqrt{5} }{2} \\\\ \mathbf{y} &=& \mathbf{\dfrac{5 \mp 2\sqrt{5} }{2}} \\ \hline \end{array}\)

 

Two pairs (x,y):
\(\begin{array}{|rcll|} \hline (x_1,~y_1) &=& \left(\dfrac{5 + 2\sqrt{5} }{2},~\dfrac{5 - 2\sqrt{5} }{2} \right) \\\\ (x_2,~y_2) &=& \left(\dfrac{5 - 2\sqrt{5} }{2},~\dfrac{5 + 2\sqrt{5} }{2} \right) \\ \hline \end{array}\)

 

\(a=5,~ b=2,~ c=5,~ d=2 \\ \mathbf{a+b+c+d = 14}\)

 

laugh

 May 31, 2021

38 Online Users

avatar
avatar
avatar
avatar
avatar
avatar