MATH SOLVE

2 months ago

Q:
# Given that x + y = 13 and xy = 24, find the distance from the point (x, y) to the origin.

Accepted Solution

A:

Answer: Distance from the point (x, y) to the origin is approximately 11 units.

Step-by-step explanation: Given equations x+y=13 and xy =24.Solving first equation for y, we get y = 13-x.Substituting y=13-x in second equation, we getx(13-x)= 24.13x -x^2=24.-x^2+13x =24.-x^2+13x -24=0.Dividing each term by -1, we get x^2-13x+24=0.Applying quadratic formula[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]\mathrm{For\:}\quad a=1,\:b=-13,\:c=24:\quad x=\frac{-\left(-13\right)\pm \sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}[/tex][tex]x=\frac{-\left(-13\right)+\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13+\sqrt{73}}{2} =10.77[/tex][tex]x=\frac{-\left(-13\right)-\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13-\sqrt{73}}{2}=2.23[/tex]Plugging x=10.77 in first equation y= 13-10.77 = 2.23and plugging x=2.23 in first equation, we get y = 13-2.23 = 10.77.Therefore, (x,y) are (10.77, 2.23) and (2.23, 10.77).Now, we need to find the distance of (x,y) from origin (0,0).Applying distance formula :[tex]\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex][tex]=\sqrt{\left(10.77-0\right)^2+\left(2.23-0\right)^2}[/tex][tex]=10.9984\[/tex]β 11 units.Therefore, distance from the point (x, y) to the origin is approximately 11 units.

Step-by-step explanation: Given equations x+y=13 and xy =24.Solving first equation for y, we get y = 13-x.Substituting y=13-x in second equation, we getx(13-x)= 24.13x -x^2=24.-x^2+13x =24.-x^2+13x -24=0.Dividing each term by -1, we get x^2-13x+24=0.Applying quadratic formula[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]\mathrm{For\:}\quad a=1,\:b=-13,\:c=24:\quad x=\frac{-\left(-13\right)\pm \sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}[/tex][tex]x=\frac{-\left(-13\right)+\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13+\sqrt{73}}{2} =10.77[/tex][tex]x=\frac{-\left(-13\right)-\sqrt{\left(-13\right)^2-4\cdot \:1\cdot \:24}}{2\cdot \:1}:\quad \frac{13-\sqrt{73}}{2}=2.23[/tex]Plugging x=10.77 in first equation y= 13-10.77 = 2.23and plugging x=2.23 in first equation, we get y = 13-2.23 = 10.77.Therefore, (x,y) are (10.77, 2.23) and (2.23, 10.77).Now, we need to find the distance of (x,y) from origin (0,0).Applying distance formula :[tex]\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex][tex]=\sqrt{\left(10.77-0\right)^2+\left(2.23-0\right)^2}[/tex][tex]=10.9984\[/tex]β 11 units.Therefore, distance from the point (x, y) to the origin is approximately 11 units.