Question 4 The sum of two numbers is 12 and the sum of the squares of these two numbers is \( 21 / 2 \) times their product. Doternine the numbers. Question 5

To solve this problem, we can set up a system of equations based on the given information. Let's denote the two numbers as \( x \) and \( y \). From the first piece of information, we have: \[ x + y = 12 \] From the second piece of information, we have: \[ x^2 + y^2 = \frac{21}{2}xy \] We can solve this system of equations to find the values of \( x \) and \( y \). Let's solve the system of equations. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=12\\x^{2}+y^{2}=\frac{21}{2}xy\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=12-y\\x^{2}+y^{2}=\frac{21}{2}xy\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(12-y\right)^{2}+y^{2}=\frac{21}{2}\left(12-y\right)y\) - step3: Simplify: \(144-24y+2y^{2}=\frac{21}{2}y\left(12-y\right)\) - step4: Move the expression to the left side: \(144-24y+2y^{2}-\frac{21}{2}y\left(12-y\right)=0\) - step5: Calculate: \(144-150y+\frac{25}{2}y^{2}=0\) - step6: Rewrite in standard form: \(\frac{25}{2}y^{2}-150y+144=0\) - step7: Multiply both sides: \(2\left(\frac{25}{2}y^{2}-150y+144\right)=2\times 0\) - step8: Calculate: \(25y^{2}-300y+288=0\) - step9: Solve using the quadratic formula: \(y=\frac{300\pm \sqrt{\left(-300\right)^{2}-4\times 25\times 288}}{2\times 25}\) - step10: Simplify the expression: \(y=\frac{300\pm \sqrt{\left(-300\right)^{2}-4\times 25\times 288}}{50}\) - step11: Simplify the expression: \(y=\frac{300\pm \sqrt{300^{2}-28800}}{50}\) - step12: Simplify the expression: \(y=\frac{300\pm 60\sqrt{17}}{50}\) - step13: Separate into possible cases: \(\begin{align}&y=\frac{300+60\sqrt{17}}{50}\\&y=\frac{300-60\sqrt{17}}{50}\end{align}\) - step14: Simplify the expression: \(\begin{align}&y=\frac{30+6\sqrt{17}}{5}\\&y=\frac{300-60\sqrt{17}}{50}\end{align}\) - step15: Simplify the expression: \(\begin{align}&y=\frac{30+6\sqrt{17}}{5}\\&y=\frac{30-6\sqrt{17}}{5}\end{align}\) - step16: Evaluate the logic: \(y=\frac{30+6\sqrt{17}}{5}\cup y=\frac{30-6\sqrt{17}}{5}\) - step17: Rearrange the terms: \(\left\{ \begin{array}{l}x=12-y\\y=\frac{30+6\sqrt{17}}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=12-y\\y=\frac{30-6\sqrt{17}}{5}\end{array}\right.\) - step18: Calculate: \(\left\{ \begin{array}{l}x=\frac{30-6\sqrt{17}}{5}\\y=\frac{30+6\sqrt{17}}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{30+6\sqrt{17}}{5}\\y=\frac{30-6\sqrt{17}}{5}\end{array}\right.\) - step19: Calculate: \(\left\{ \begin{array}{l}x=\frac{30+6\sqrt{17}}{5}\\y=\frac{30-6\sqrt{17}}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{30-6\sqrt{17}}{5}\\y=\frac{30+6\sqrt{17}}{5}\end{array}\right.\) - step20: Check the solution: \(\left\{ \begin{array}{l}x=\frac{30+6\sqrt{17}}{5}\\y=\frac{30-6\sqrt{17}}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{30-6\sqrt{17}}{5}\\y=\frac{30+6\sqrt{17}}{5}\end{array}\right.\) - step21: Rewrite: \(\left(x,y\right) = \left(\frac{30+6\sqrt{17}}{5},\frac{30-6\sqrt{17}}{5}\right)\cup \left(x,y\right) = \left(\frac{30-6\sqrt{17}}{5},\frac{30+6\sqrt{17}}{5}\right)\) The solution to the system of equations is: \[ (x, y) = \left(\frac{30+6\sqrt{17}}{5}, \frac{30-6\sqrt{17}}{5}\right) \text{ or } (x, y) = \left(\frac{30-6\sqrt{17}}{5}, \frac{30+6\sqrt{17}}{5}\right) \] Therefore, the two numbers are \( \frac{30+6\sqrt{17}}{5} \) and \( \frac{30-6\sqrt{17}}{5} \), or \( \frac{30-6\sqrt{17}}{5} \) and \( \frac{30+6\sqrt{17}}{5} \).