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Find all real numbers

The question is, find all real numbers x, y and z which satisfy the simulataneous equations

x2−4y+7=0,y2−6z+14=0 and z2−2x−7=0 x^2 - 4y + 7 = 0, y^2 - 6z + 14 = 0 \ \text{and} \ z^2 - 2x - 7 = 0

I got x = 1, y = 2 and z = 3, are those the only solutions?

Thanks.
(edited 10 years ago)
Original post by 0x2a
The question is, find all real numbers x, y and z which satisfy the simulataneous equations

x2−4y+7=0,y2−6z+14=0 and z2−2x−7=0 x^2 - 4y + 7 = 0, y^2 - 6z + 14 = 0 \ \text{and} \ z^2 - 2x - 7 = 0

I got x = 1, y = 2 and z = 3, are those the only solutions?

Thanks.


http://www.wolframalpha.com/input/?i=solve+x%5E2-4y%2B7%3D0+and+y%5E2-6z%2B14%3D0+and+z%5E2-2x-7%3D0+over+the+reals
Reply 2
Avatar for 0x2a
0x2a
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2
Original post by Mr M

I forgot the ultimate cheating machine.

Thank you.
Reply 3
Avatar for Hasufel
Hasufel
16
Yep.

add all of these equations to each other, and factorise (completing the square) and you get:

(x−1)2+(y−2)2+(z−3)2=0(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=0 (a single point)
(edited 10 years ago)
Original post by Hasufel
Yep.

add all of these equations to each other, and factorise (completing the square) and you get:

(x−1)2+(y−2)2+(z−3)2=0(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=0 (a single point)

Is that a sphere then?
(edited 10 years ago)
Reply 5
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0x2a
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Original post by keromedic
Is that a sphere then?

The equation is in the form which gives a sphere, but it's radius is 0.
Original post by 0x2a
The equation is in the form which gives a sphere, but it's radius is 0.

Does that make it a sphere in the complex plane?
Reply 7
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0x2a
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2
Original post by keromedic
Does that make it a sphere in the complex plane?

The z is only used here as another "axis" along with the x and the y. A shape of the form (x−x1)2+(y−y1)2+(z−z1)2=r2 (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = r^2 would appear in R3 \mathbb{R}^3 , unless you're explicitly dealing with complex numbers.
(edited 10 years ago)
Original post by keromedic
Does that make it a sphere in the complex plane?

It makes it a point.

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