# Find the coordinates of the points of intersection of the two curves. (y=x^2-1 & y=6/x^2)?

Apr 17, 2018

$\left(- \sqrt{3} , 2\right) \text{ , } \left(\sqrt{3} , 2\right)$

#### Explanation:

$y = {x}^{2} - 1$

$y = \frac{6}{x} ^ 2$

by solving the two equations simultaneously

${x}^{2} - 1 = \frac{6}{x} ^ 2$

${x}^{4} - {x}^{2} - 6 = 0$

Factorize

$\left({x}^{2} - 3\right) \left({x}^{2} + 2\right) = 0$

${x}^{2} = 3$ $\text{ , }$color(red)(x^2=-2 " refused as it's solution is not real"

$x = \pm \sqrt{3}$

Substitute in any of the equations to find the $y$ coordinates of the points of intersection

$f \left(- \sqrt{3}\right) = 2$$\text{ , }$$f \left(\sqrt{3}\right) = 2$

so the two points of intersection will be:

$\left(- \sqrt{3} , 2\right) \text{ , } \left(\sqrt{3} , 2\right)$