Solve Quadratic Equation: √(5x-1) - √x=1

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Discussion Overview

The discussion revolves around solving the quadratic equation √(5x-1) - √x=1. Participants explore the implications of squaring both sides of the equation and the necessity of verifying potential solutions against the original equation.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution process leading to x=1/4 and x=1, questioning why x=1/4 is not a valid root.
  • Another participant clarifies that substituting x=1/4 into the original equation results in 0, indicating it does not satisfy the equation, thus it is not a solution.
  • There is a discussion about the general principle that squaring both sides of an equation can introduce extraneous solutions, particularly in equations involving radicals.
  • Some participants suggest that this issue is not limited to equations with radicals, as similar problems can arise from operations that yield multiple results.

Areas of Agreement / Disagreement

Participants generally agree on the need to verify solutions after squaring both sides of an equation, but there is some debate regarding the extent to which this issue applies to different types of equations.

Contextual Notes

The discussion highlights the importance of checking solutions in the context of operations that can introduce extraneous roots, particularly when squaring both sides of an equation.

azizlwl
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solve √(5x-1) - √x=1 eq. 1
transpose one of the radicals: √(5x-1)=√x+1
Square: 5x-1=x+2√x+1
Collect terms: 4x-2=2√x or 2x-1=√x
Square: 4x^2-4x+1=x, x=1/4 , 1.

x=1/4 is not the root by substitution to equation 1.
...

My question what makes x =1/4 is not the root, only x=1 is the root.
 
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The fact that \sqrt{5(1/4)- 1}= \sqrt{5/4- 1}= \sqrt{1/4}= 1/2
while \sqrt{1/4}= 1/2 also so that if x= 1/4 \sqrt{5x-1}- \sqrt{x}= 0, not 1.

That is, it is not a solution because it does not satisfy the equation!

Any time we square both sides of an equation, or multiply both sides of an equation by something involving "x", we may introduce "new" solutions that satisfy the new equation but not the original one.
 
Does this occurs only for equations with radicals since normally we do not test the values.
 
azizlwl said:
Does this occurs only for equations with radicals since normally we do not test the values.



This may occur anytime we're trying to solve equations by methods that involve some operation that can

give two or more different results. In this case, when you square stuff, the difference between positive and

negative terms disappears and thus it is necessary, at the end of the process, to check each and every

supposed solution in the original equation.

DonAntonio
 

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