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

In summary: The claim made is correct and only positive solutions for the square root of energy are physically acceptable.In summary, the conversation discusses a quadratic equation where the variable to be found is the square root of energy. The participants question whether the claim that only positive solutions for the square root of energy are physically acceptable is correct. It is determined that the claim is indeed correct as negative solutions would be extraneous. Additionally, it is mentioned that this equation may be solving for a scaled version of velocity, in which case the square of the negative root would give the energy.
  • #1
Alex_physics
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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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  • #3
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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