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How many real solutions are there to the equation x^2 =1/(x+3)

 Apr 14, 2023
 #1
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There are no real solutions to the equation x^2 = 1/(x+3).

To see this, we can try to solve for x by isolating x on one side of the equation. Multiplying both sides of the equation by x+3, we get x^2(x+3) = 1. Expanding the left-hand side, we get x^3 + 3x^2 = 1.

This equation is a polynomial equation of degree 3. Polynomial equations of degree 3 have at most 3 roots, which can be real or complex numbers. In this case, the roots of the equation x^3 + 3x^2 - 1 = 0 are complex numbers. Therefore, there are no real solutions to the equation x^2 = 1/(x+3).

 Apr 14, 2023
 #2
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Complex roots appear in pairs, as complex conjugates.

A cubic equation will have three real roots or, a single real root plus a conjugate pair.

Guest Apr 14, 2023

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