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

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Answer 1 · 2018-04-17T10:11:56

$(- \sqrt{3}, 2), (\sqrt{3}, 2)$ $(-sqrt3,2) " , "(sqrt3,2)$

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

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

by solving the two equations simultaneously

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

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

Factorize

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

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

$x = \pm \sqrt{3}$ $x=+-sqrt3$

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

$f (- \sqrt{3}) = 2$ $f(-sqrt3)=2$ $,$ $" , "$ $f (\sqrt{3}) = 2$ $f(sqrt3)=2$

so the two points of intersection will be:

$(- \sqrt{3}, 2), (\sqrt{3}, 2)$ $(-sqrt3,2) " , "(sqrt3,2)$

I hope this was helpful.