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# Algebra

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Find all solutions to the system
a + b = 14
a^3 + b^3 = 812 + 3ab

Jul 12, 2021

#1
+26222
+2

Find all solutions to the system
$$a + b = 14\\ a^3 + b^3 = 812 + 3ab$$

$$\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a+b = 14 \\ 14^3 &=& a^3+b^3 +3*14ab \quad | \quad a^3+b^3 = 812+3ab \\ 14^3 &=& 812+3ab +3*14ab \\ 14^3 &=& 812+3*15ab \\ 3*15ab &=& 14^3- 812 \\ 3*15ab &=& 1932 \quad | \quad : 3 \\ 15ab &=& 644 \quad | \quad b=14-a \\ 15a(14-a) &=& 644 \\ a(14-a) &=& \dfrac{644}{15} \\ 14a-a^2 &=& \dfrac{644}{15} \\ \mathbf{a^2-14a+ \dfrac{644}{15}} &=& \mathbf{0} \\ \hline a &=& \dfrac{ 14\pm \sqrt{14^2 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm \sqrt{4*49 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm 2\sqrt{49 -\dfrac{644}{15}} }{2} \\ \\ a &=& 7\pm \sqrt{49 -\dfrac{644}{15}} \\ \\ a &=& 7\pm \sqrt{\dfrac{735-644}{15}} \\ \\ \mathbf{a} &=& \mathbf{7\pm \sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

1.

$$\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7+\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

2.

$$\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7-\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

Jul 12, 2021

#1
+26222
+2

Find all solutions to the system
$$a + b = 14\\ a^3 + b^3 = 812 + 3ab$$

$$\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a+b = 14 \\ 14^3 &=& a^3+b^3 +3*14ab \quad | \quad a^3+b^3 = 812+3ab \\ 14^3 &=& 812+3ab +3*14ab \\ 14^3 &=& 812+3*15ab \\ 3*15ab &=& 14^3- 812 \\ 3*15ab &=& 1932 \quad | \quad : 3 \\ 15ab &=& 644 \quad | \quad b=14-a \\ 15a(14-a) &=& 644 \\ a(14-a) &=& \dfrac{644}{15} \\ 14a-a^2 &=& \dfrac{644}{15} \\ \mathbf{a^2-14a+ \dfrac{644}{15}} &=& \mathbf{0} \\ \hline a &=& \dfrac{ 14\pm \sqrt{14^2 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm \sqrt{4*49 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm 2\sqrt{49 -\dfrac{644}{15}} }{2} \\ \\ a &=& 7\pm \sqrt{49 -\dfrac{644}{15}} \\ \\ a &=& 7\pm \sqrt{\dfrac{735-644}{15}} \\ \\ \mathbf{a} &=& \mathbf{7\pm \sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

1.

$$\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7+\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

2.

$$\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7-\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ \hline \end{array}$$

heureka Jul 12, 2021