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When the sum of 8 and twice a positive number is subtracted from the square root of the number 0 results?
Wiki User
∙ 8y ago
8
Cassandra Carroll
Wiki User
∙ 8y agoThere can be no possible solution.
Suppose the number is x where x>0, then
then twice that number is 2x
and the sum of 8 and twice that number is 8 + 2x
This value, subtracted from the square root of the number is sqrt(x) - (8 + 2x)
Therefore, sqrt(x) - (8 + 2x) = 0
adding 8 + 2x to both sides: sqrt(x) = 8 + 2x
squaring both sides: x = 64 + 32x + 4x^2
swapping sides and simplifying: 4x^2 + 31x + 64 = 0
The determinant for this quadratic is 31^2 - 4*4*64 = 961 - 1024 = -63.
There is therefore no real solution, only complex solutions.
Therefore there cannot be a positive number x.
Wiki User
∙ 8y agoThis doesn't work as written. There is no positive number solution for x. If that was square instead of square root, you could use 4.
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