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# Vieta's Formula

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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

#1
+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 $$(x_1,x_2)$$.

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

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

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

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

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

Jun 4, 2019

#1
+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 $$(x_1,x_2)$$.

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

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

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

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

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

heureka Jun 4, 2019