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The graphs of $y=x^2$ and $x+y=0$ intersect at two points. Find the distance between these two points.

User Selman
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1 Answer

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Answer:

To find the points of intersection, we can set the equations of the graphs equal to each other:

$x^2 = -x$

Rearranging the terms, we have:

$x^2 + x = 0$

Factoring out an $x$, we get:

$x(x + 1) = 0$

This equation is satisfied when $x = 0$ or $x = -1$.

Substituting these values back into the equation $y = x^2$, we find the corresponding $y$-coordinates.

For $x = 0$, we have $y = (0)^2 = 0$.

For $x = -1$, we have $y = (-1)^2 = 1$.

Therefore, the two points of intersection are $(0, 0)$ and $(-1, 1)$.

To find the distance between these two points, we can use the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Using the coordinates of the two points, we have:

$\text{Distance} = \sqrt{((-1) - 0)^2 + (1 - 0)^2}$

Simplifying,

$\text{Distance} = \sqrt{1 + 1}$

$\text{Distance} = \sqrt{2}$

Therefore, the distance between the two points of intersection is $\sqrt{2}$.

User Bigtv
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