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# What is the distance between the two intersections of $y=x^2$ and $x+y=1$?

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What is the distance between the two intersections of $y=x^2$ and $x+y=1$?

Feb 26, 2020

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What is the distance between the two intersections of $$y=x^2$$ and $$x+y=1$$?

$$\begin{array}{|lrcll|} \hline & \text{intersections:} \\ \hline (1) & y &=& x^2 \\ \\ (2) & x+y &=& 1 \quad | \quad y = x^2 \\ & x+x^2 &=& 1 \\ & \mathbf{x^2+x-1} &=& \mathbf{0} \\ & x &=& \frac{-1\pm \sqrt{1-4(-1)} }{2} \\ & x &=& \frac{-1\pm\sqrt{5} }{2} \\ & \mathbf{x_1 = \dfrac{-1+\sqrt{5} }{2}} && \mathbf{x_2 = \dfrac{-1-\sqrt{5} }{2}} \\ \hline \end{array}$$

$$\begin{array}{|rcl|rcl|} \hline x_1-x_2 &=& \dfrac{-1+\sqrt{5} }{2}-\dfrac{-1-\sqrt{5} }{2} & x_1+x_2 &=& \dfrac{-1+\sqrt{5} }{2}+\dfrac{-1-\sqrt{5} }{2} \\\\ &=& -\frac{1}{2} +\frac{\sqrt{5} }{2}+\frac{1}{2}+\frac{\sqrt{5} }{2} & &=& -\frac{1}{2} +\frac{\sqrt{5} }{2}-\frac{1}{2}-\frac{\sqrt{5} }{2} \\\\ \mathbf{x_1-x_2} &=& \mathbf{\sqrt{5}} &\mathbf{x_1+x_2} &=& \mathbf{-1} \\ \mathbf{\left(x_1-x_2\right)^2} &=& \mathbf{5} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline y_1-y_2 &=& x_1^2-x_2^2 \\ &=& (x_1 -x_2)(x_1 +x_2) \\ &=& \sqrt{5}(-1) \\ \mathbf{y_1-y_2} &=& \mathbf{-\sqrt{5}} \\ \mathbf{\left(y_1-y_2\right)^2} &=& \mathbf{5} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{The distance} \\ \hline &=& \sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} \\ &=& \sqrt{5+5} \\ &=& \sqrt{10} \\ &=&\mathbf{ 3.16227766017\ldots} \\ \hline \end{array}$$

Feb 26, 2020