Why no solutions exist
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# Why no solutions exist

[From: ] [author: ] [Date: 11-05-19] [Hit: ]
0);(0,-1);(-1,The equation represents a circle with radius 1-1 = x^2 + y^2 this represent circle of radius 1 & center (0,so any x,......
-1 = sqrt(x^2 +y^2)

dont understand why

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x² is always positive..
y² is also always positive..
sp √(x²+y²) is always positive..

SO it can't equal -1

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When you have √, ALWAYS take principal square root (i.e. positive square root)
So left side = -1 < 0, but right side >= 0

Therefore there is no solution.

====================

Please note that in quadratic formula: x = (-b ± √(b²-4ac)) / (2a)
we EXPLICITLY use ± in front of square root, since there is a difference
between √(b²-4ac) and -√(b²-4ac)

Example:
√64 = 8
-√64 = -8

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actually solutions exist
on squaring both sides
1 = x^2 +y^2

The solutions are (0,1);(1,0);(0,-1);(-1,0)

The equation represents a circle with radius 1

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1 = x^2 + y^2 this represent circle of radius 1 & center (0,0) on xy plane
so any x,y value on circle is a solution for equation

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a solution does exist

x = 1
y = 0

whoever told you no solutions exist is wrong

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thanks
1
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