Solving Nice Exponential Equation | Solve x=x^2+6x^1/2. | Exponential Equation | Algebra.

Solving Nice Exponential Equation x=x^2+6x^1/2. Solving Exponential or radical Equation Systematically is one of the best video on algebraic simplification you will ever enjoy watching as it is loaded with so many insights and tricks into solving exponential equations of this kind. #algebra #exponentialequations #olympiadmathematics #factorization #simplification

In this video, you will learn how to introduce another alphabet to replace the square root from the expression and simplify systematically according to the algebraic terms and signs involve.

You will learn how to form a quartic equation from the radical equation above and how to factorize the quartic equation completing. Here, you will learn how to use the trial by error method to solve polynomial equations also.
You will also learn how to create a quartic equation from the above exponential equation the easy way. I will show you how to solve the resulted quadratic equation by using the factorization method in order to get x variable.

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[asparkdeity]

My initial thought is to make the expression into a quartic expression with respect to x^1/2 … now I’ll watch the video :D

After watching: great solution! I was just wondering however, if x = 4 is a solution because then x^1/2 is 2 rather than -2 (and we assume positive root), I don’t think there are any solutions, real or complex, to the equation sqrt(x) = a for any a < 0. Thank you very much for this video!

[sy]

Thank you for explaining. 　I checked the four solutions in this video. And I got to know "4" cannot be the solution.
(left side) = x = 4 (right side) = x^2+6x^(1/2) = 4^2+6×4^(1/2) = 16+6×2 = 16+12 = 28 Therefore, 4 should be rejected.
・・・・ Or am I misunderstand something? ・・・・

[letstalkwithbigsheila]

This is good. We need more of this. Just maybe it will make many like math.

[innos.]

The substitution approach was a good one i.e. the part where we let y = x^½

[Gaby-lzdf]

❗❗❗ The solution was generally good. Thank you. But it had to be notice immediatelly that it is an impossibility that x^½ = - 2 < 0 because always x^½ >= 0 by the definition of the radical. By squaring up are introduced invalid solutions. So x = 4 is not a solution.
In conclusion only the other 3 solutions are valid. 🙂