1. For what real values of does the quadratic have nonreal roots? Enter your answer as an interval.
2. Find one pair $(x,y)$ of real numbers such that $x + y = 4$ and $x^3 + y^3 = 100.$
I honestly don't understand how to solve these, if someone could provide an explanation w/an answer that would help a lot.
+26367 2.
Find one pair \((x,y)\) of real numbers such that \(x + y = 4\) and \(x^3 + y^3 = 100\).
\(\begin{array}{|rcll|} \hline x^3+y^3 &=& (x+y)(x^2-xy+y^2 ) \quad & | \quad x + y = 4,\ x^3 + y^3 = 100 \\ 100 &=& 4(x^2-xy+y^2 ) \quad & | \quad :4 \\ 25 &=& x^2-xy+y^2 \\ \mathbf{x^2+y^2 -xy} &=& \mathbf{25} \qquad (1) \\\\ (x+y)^2 &=& 16 \\ x^2+2xy + y^2 &= 16 \\ \mathbf{x^2+y^2 } &=& \mathbf{16-2xy} \qquad (2) \\ \hline x^2+y^2 -xy &=& 25 \quad & | \quad x^2+y^2 = 16-2xy \\ 16-2xy -xy &=& 25 \\ 16-3xy &=& 25 \\ \ldots \\ \mathbf{xy} &=& \mathbf{-3} \qquad (3) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (I): & xy &=& -3 \\ & y &=& \dfrac{-3}{x} \\\\ (II): & x + y &=& 4 \quad & | \quad y = \dfrac{-3}{x} \\ & x + \dfrac{-3}{x} &=& 4 \quad & | \quad * x \\ & x^2 -3 &=& 4x \\ & x^2 -4x-3 &=& 0 \\\\ & x &=& \dfrac{4\pm \sqrt{16-4(-3)} } {2} \\ & x &=& \dfrac{4\pm \sqrt{28} } {2} \\ & x &=& \dfrac{4\pm \sqrt{4*7} } {2} \\ & x &=& \dfrac{4\pm 2\sqrt{ 7} } {2} \\ & x &=& 2\pm \sqrt{7} \\\\ & \mathbf{x_1} &=& \mathbf{2+\sqrt{7}} \\ & x_1+y_1 &=& 4 \quad & \quad x_1 = 2+\sqrt{7} \\ & 2+\sqrt{7} + y_1 &=& 4 \\ & \mathbf{y_1} &=& \mathbf{2-\sqrt{7}} \\\\ & \mathbf{x_2} &=& \mathbf{2-\sqrt{7}} \\ & x_2+y_2 &=& 4 \quad & \quad x_2 = 2-\sqrt{7} \\ & 2-\sqrt{7} + y_2 &=& 4 \\ & \mathbf{y_2} &=& \mathbf{2+\sqrt{7}} \\ \hline \end{array}\)
\((x,y)=(\mathbf{2+\sqrt{7}},\ \mathbf{2-\sqrt{7}})~ \text{or}~ \\ (x,y)=(\mathbf{2-\sqrt{7}},\ \mathbf{2+\sqrt{7}})\)
