# algebra

Find the real solutions

8x^2 + y^2 = 25
8X^2 - y^2 = 39

Is this no solution because I tried solving it and i got the square root of 32 for y, but when i plugged it in to get x there was a negative under the square root sign which means its no solution, right? please help...thanks!

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1. The first one is an ellipse
The second one is a hyperbola
What do you mean by solutions?
Do you want values of x where y = 0 or what?

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posted by Damon
2. Oh, I see from your earlier questions, solve the intersections of ellipse and hyperbola.
Well both are centered on the origin, so the hyperbola either cuts the ellipse or misses it entirely.

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posted by Damon
3. okay thanks! so that means...there is no solution for solving it algebraically?

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posted by meg
4. The ellipse is from x = -2 to x = +2 and y = -1 to y = + 1
The hyperbola is above y = 5 and below y = -5

THEREFORE THEY NEVER HIT

Let's say from the ellipse y^2 = (25 - 8 x^2)
then from the hyperbola
8 x^2 - (25 - 8 x^2) = 39
16 x^2 -25 = 39
16 x^2 = 64
x^2 = 4
x = 2, x = -2
y^2 = 25 - 8(4) = 25 - 32 = imaginary y
This is at the very ends of the ellipse, but the hyperbola does not go there

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posted by Damon
5. whoa, too quick
the ellipse is from x = -5/(2 sqrt 2) to +5/(2 sqrt 2) which is about -1.77 to +1.77 and y from -5 to +5
The hyperbola is left of - 2.2 and right of +2.2
They do not hit no matter how you look at them :)

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posted by Damon

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