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

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SUMMARY

The discussion centers on the acceptability of negative roots in a quadratic equation involving the square root of energy, specifically in the form of (a*sqrt(E)^2 + b*sqrt(E) + c = 0). It is established that while energy (E) is real and positive, the square root of energy (sqrt(E)) can yield negative solutions. However, only positive roots are physically acceptable when considering kinetic energy, as negative roots would be extraneous. The conversation also touches on the interpretation of the equation as potentially representing a scaled version of velocity, where negative values for velocity are permissible.

PREREQUISITES
  • Understanding of quadratic equations and their solutions
  • Familiarity with the concept of kinetic energy, specifically E = 1/2 mv^2
  • Knowledge of the properties of square roots and their implications in physics
  • Basic grasp of the relationship between energy and velocity in physical systems
NEXT STEPS
  • Research the implications of negative roots in quadratic equations
  • Explore the derivation and applications of kinetic energy equations
  • Study the mathematical properties of square roots in physical contexts
  • Investigate the relationship between velocity and energy in classical mechanics
USEFUL FOR

This discussion is beneficial for physics students, educators, and professionals involved in mechanics, particularly those focusing on energy concepts and quadratic equations in physical applications.

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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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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