Get the answer to "Two equations and their graphs are given. Find..."

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Answered question

2021-09-10

Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system

$x^{2} + y = 0, x^{3} - 2 x - y = 0$

SchulzD

Skilled2021-09-11Added 83 answers

Given equations are:

$x^{2} + y = 0, x^{3} - 2 x - y = 0$

$x^{2} + y = 0 \Rightarrow y = - x^{2} . . . (1)$

$x^{3} - 2 x - y = 0 . . . (2)$

Substituting $y = - x^{2}$ from equation (1) in equation (2) we get:

$x^{3} - 2 x - y = 0$

$\Rightarrow x^{3} - 2 x - (- x^{2}) = 0$

$\Rightarrow x^{3} - 2 x + x^{2} = 0$

$\Rightarrow x^{3} + x^{2} - 2 x = 0$

$\Rightarrow x (x^{2} + x - 2) = 0$
$\Rightarrow x (x^{2} + 2 x - x - 2) = 0$ $\Rightarrow x (x (x + 2) - (x + 2)) = 0$

$\Rightarrow x (x + 2) (x - 1) = 0$

$\Rightarrow x = 0$ ,or, $x = - 2$ ,or, $x = 1.$

Substituting, $x = 0$ ,in equation (1) we get y as: $y = - 2 = 0, (0, 0)$ .

Substituting, $x = - 2$ ,in equation (1) we get y as: $y = - 22 = - 4, (- 2, - 4)$ .

Substituting, $x = 1$ ,in equation (1) we get y as: $y = - 12 = - 1, (1, - 1)$ .

Intersection points: $(- 2, 4), (0, 0), (1, - 1)$

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