+0  
 
0
243
1
avatar+1207 

The equations \(x^3 + 5x^2 + px + q = 0\) and \(x^3 + 7x^2 + px + r = 0\) have two roots in common. If the third root of each equation is represented by \(x_1\) and \(x_2\) respectively, compute the ordered pair \((x_1,x_2)\).

 Jun 4, 2019

Best Answer 

 #1
avatar+24388 
+2

Vieta's Formula
The equations  \(x^3 + 5x^2 + px + q = 0\) and \(x^3 + 7x^2 + px + r = 0\) have two roots in common.
If the third root of each equation is represented by \(x_1\) and \(x_2\) respectively,
compute the ordered pair \((x_1,x_2)\).

 

\(\begin{array}{|rcll|} \hline x^3 + 5x^2 + px + q &=& 0,\ \text{ the roots are }~ x_1,x_3,x_4. \\ x^3 + 7x^2 + px + r &=& 0,\ \text{ the roots are }~ x_2,x_3,x_4. \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \\ 5+x_1 &=& -(x_3+x_4) \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \\ 7+x_2 &=& -(x_3+x_4) \\\\ 5+x_1 &=& 7+x_2 \\ \mathbf{x_1-x_2} &=& \mathbf{2} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline x_1x_3+x_1x_4+x_3x_4 = p &=& x_2x_3+x_2x_4+x_3x_4 \\ x_1x_3+x_1x_4 &=& x_2x_3+x_2x_4 \\ x_1(x_3+x_4)-x_2(x_3+x_4) &=& 0 \\ (x_3+x_4)(x_1-x_2) &=& 0 \quad | \quad \mathbf{x_1-x_2 = 2} \\ (x_3+x_4)*2 &=& 0 \quad | \quad : 2 \\ x_3+x_4 &=& 0 \\ \mathbf{x_3} &=& \mathbf{-x_4} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 5 &=& -(x_1-x_4+x_4) \\ 5 &=& -x_1 \\ \mathbf{x_1} &=& \mathbf{-5} \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 7 &=& -(x_2-x_4+x_4) \\ 7 &=& -x_2 \\ \mathbf{x_2} &=& \mathbf{-7} \\ \hline \end{array}\)

 

\((x_1,~x_2) = (-5,~-7)\)

 

laugh

 Jun 4, 2019
 #1
avatar+24388 
+2
Best Answer

Vieta's Formula
The equations  \(x^3 + 5x^2 + px + q = 0\) and \(x^3 + 7x^2 + px + r = 0\) have two roots in common.
If the third root of each equation is represented by \(x_1\) and \(x_2\) respectively,
compute the ordered pair \((x_1,x_2)\).

 

\(\begin{array}{|rcll|} \hline x^3 + 5x^2 + px + q &=& 0,\ \text{ the roots are }~ x_1,x_3,x_4. \\ x^3 + 7x^2 + px + r &=& 0,\ \text{ the roots are }~ x_2,x_3,x_4. \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \\ 5+x_1 &=& -(x_3+x_4) \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \\ 7+x_2 &=& -(x_3+x_4) \\\\ 5+x_1 &=& 7+x_2 \\ \mathbf{x_1-x_2} &=& \mathbf{2} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline x_1x_3+x_1x_4+x_3x_4 = p &=& x_2x_3+x_2x_4+x_3x_4 \\ x_1x_3+x_1x_4 &=& x_2x_3+x_2x_4 \\ x_1(x_3+x_4)-x_2(x_3+x_4) &=& 0 \\ (x_3+x_4)(x_1-x_2) &=& 0 \quad | \quad \mathbf{x_1-x_2 = 2} \\ (x_3+x_4)*2 &=& 0 \quad | \quad : 2 \\ x_3+x_4 &=& 0 \\ \mathbf{x_3} &=& \mathbf{-x_4} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{5} &=& \mathbf{-(x_1+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 5 &=& -(x_1-x_4+x_4) \\ 5 &=& -x_1 \\ \mathbf{x_1} &=& \mathbf{-5} \\\\ \mathbf{7} &=& \mathbf{-(x_2+x_3+x_4)} \quad | \quad \mathbf{x_3 = -x_4} \\ 7 &=& -(x_2-x_4+x_4) \\ 7 &=& -x_2 \\ \mathbf{x_2} &=& \mathbf{-7} \\ \hline \end{array}\)

 

\((x_1,~x_2) = (-5,~-7)\)

 

laugh

heureka Jun 4, 2019

69 Online Users

avatar
avatar
avatar