Question

I have a practice question that I need explained and answered. Thank you - Rose

Answer 1

To determine the x - coordinate of the distance between two points:

`d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

The distance between the two points is estimated using the above formular

`\begin{gathered} x_2=4 \\ x_1=?_{} \\ y_1=-1 \\ y_2=9 \\ d=6\sqrt[]{6} \end{gathered}`
`\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ 6\sqrt[]{6}=\sqrt[]{(x-4)^2+(9--1)^2} \\ \text{square both side } \\ (6\sqrt[]{6)^2}=(\sqrt[]{(x-4)^2+10^2})^2 \\ 216=(x-4)^2+100 \\ 216-100=(x-4)^2 \end{gathered}`
`\begin{gathered} 116=(x-4)(x-4) \\ 116=x^2-8x+16 \\ 100=x^2-8x \\ x^2-8x-100=0 \end{gathered}`

Solve using quadratic formular

`\frac{-b\pm\sqrt[]{b^2}-4ac}{2a}`
`\begin{gathered} \frac{-b\pm\sqrt[]{b^2}-4ac}{2a}\ldots..\text{ a= 1 , b = -8 , c = -100} \\ \frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(-100)}}{2(1)} \\ \frac{8\pm\sqrt[]{64+400}}{2} \\ \frac{8\pm\sqrt[]{464}}{2}=\frac{8\pm4\sqrt[]{29}}{2} \\ \frac{2(4\pm2\sqrt[]{29)}}{2}=4\pm2\sqrt[]{29} \end{gathered}`

Therefore the correct answer for the x - coordinates are:

`\begin{gathered} x=4+2\sqrt[]{29}\text{ and } \\ x_{}=4-2\sqrt[]{29} \end{gathered}`