B Can negative roots of a quadratic equation for √E be physically acceptable?

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The discussion centers on the acceptability of negative roots in a quadratic equation where the variable is the square root of energy. It is argued that while energy (E) is real and positive, negative solutions for sqrt(E) could yield a positive E, suggesting their validity. However, if the equation specifically pertains to kinetic energy, only positive roots should be considered acceptable, as negative solutions would be extraneous. The conversation also touches on the possibility that the equation might represent a scaled version of velocity, where negative roots could still be relevant. Ultimately, the conclusion is that the equation is indeed for the square root of energy, reinforcing the need for positive roots in this context.
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In a certain type of problem, a quadratic equation is formed with the square root of energy being the variable to be found ex: (a*sqrt(E)^2+b*sqrt(E)+c=0). Then they claim since energy (E) is real and positive, only solutions to the quadratic equation in sqrt(E ) being real and positive are physically acceptable.

sqrt(E) can still be negative (and so the solutions to the quadratic equation) which gives a positive E is my assumption. Is their claim correct ?
 
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In the equation they give, it is possible that they are actually solving for a scaled version of the velocity. If the equation is truly for the square root of the energy,(I'm presuming their referring to kinetic energy and it doesn't include potential energy which can be negative), then only positive roots would be acceptable and negative solutions would be extraneous. If the equation is solving for a scaled version of the velocity, (oftentimes energy ## E=\frac{1}{2}mv^2 ##), then taking the square of the negative root would give the energy. (The velocity ## v ## is allowed to be negative.)
 
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Charles Link said:
In the equation they give, it is possible that they are actually solving for a scaled version of the velocity. If the equation is truly for the square root of the energy,(I'm presuming their referring to kinetic energy and it doesn't include potential energy which can be negative), then only positive roots would be acceptable and negative solutions would be extraneous. If the equation is solving for a scaled version of the velocity, (oftentimes energy ## E=\frac{1}{2}mv^2 ##), then taking the square of the negative root would give the energy. (The velocity ## v ## is allowed to be negative.)

Thank you for your answer.
The equation is truly for the square root of energy.
 
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