MHB Solving Equation: x = sqrt(3x + x^2 - 3sqrt(3x + x^2))

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The equation x = sqrt(3x + x^2 - 3sqrt(3x + x^2) is solved by first squaring both sides, leading to x^2 = 3x + x^2 - 3sqrt(3x + x^2). Simplifying this results in -3x = -3sqrt(3x + x^2), which further simplifies to x = sqrt(3x + x^2). Squaring again gives x^2 = 3x + x^2, leading to the conclusion that 0 = 3x. The only real solution is x = 0, confirming the correctness of the solution process.
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Find all real solutions of the equation.

$x = \sqrt{3x + x^2 - 3\sqrt{3x + x^2}}$

Must I square each side twice to start?

$(x)^2 = [\sqrt{3x + x^2 - 3\sqrt{3x + x^2}}]^2$

$x^2 = 3x + x^2 - 3\sqrt{3x + x^2}$

$x^2 - x^2 - 3x = -3\sqrt{3x + x^2}$

$-3x = -3\sqrt{3x + x^2}$

$x = \sqrt{3x + x^2}$

$(x)^2 = [\sqrt{3x + x^2}]^2$

$x^2 = 3x + x^2$

$x^2 - x^2 = 3x$

$0 = 3x$

$0/3 = x$

$0 = x$

Is this right?
 
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Yes, it's correct.
 
Good to know that I am right.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

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