A Nice Algebra Problem | Math Olympiad | Find all x?

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Find the value of x?
How to solve x² - 6 = √(x + 6)
In this video, we'll show you How to Solve Math Olympiad Question A Nice Algebra Problem x² - 6 = √(x + 6) in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #maths #math #algebra

[@DenisMathSolver]

Good strategies and procedures

[@SALogics]

Thank you so much for liking! ❤

[@xystusojadi8990]

I tried to solve before watching a couldnt think of that Great strat

[@1234larry1]

The reason -2 doesn’t work as a factor, even though it appears as a factor in the quartic, is when you substitute it in the original equation, you get -2=sqrt(4) and -2 is not the principle sqrt of 4.

[@SALogics]

Very nice! ❤

[@changliu3580]

Let y=√(x+6), we have: x^2-6=y, y^2-6=x, subtract the latter from the former, x^2-y^2=y-x, therefore: x=y, or x+y=-1, from the first, we get x=3 to be a valid solution, from the 2nd, we get (-1-√21)/2

[@SALogics]

Very nice! ❤

[@princemensah9731]

Please how do you come up with these factors and expansion of terms. Example from -3x=2x-5x. Is it a theorem or law you are applying? Thanks

[@SALogics]

Just by using the factors and divisibility of numbers included in the expression! ❤

[@bonowin]

I asked chat-gpt4.0 three times, but he couldn't give me the answer. What did you do in extracting the common factors? I couldn't understand.

[@SALogics]

By findind the common factor of the numbers in the expression! ❤

[@alucardthespy5539]

Automatically, I want to plug in values of x so x + 6 produces a square... then we'll check them with x^2 - 6 x + 6 = 1 x = -7 x^2 - 6 = 49 - 6 = 43 ≠ √1 x + 6 = 4 x = -2 x^2 - 6 = 4 - 6 = -2 = √4 x + 6 = 9 x = 3 x^2 - 6 = 9 - 6 = 3 = √9 x + 6 = 16 x = 10 x^2 - 6 = 100 - 6 = 94 ≠ √16 So, already I found two solutions... x = -2, and x = 3, now let's find the other solutions... x^2 - 6 = √(x + 6) Square both sides... x^4 - 12x^2 + 36 = x + 6 Move everything to the left x^4 - 12x^2 - x + 30 = 0 x = -2, (x + 2) x = 3, (x - 3) (x + 2)(x - 3) = x^2 - x - 6 Now divide the polynomial by (x^2 - x - 6) (x^4 - 12x^2 - x + 30)/(x^2 - x - 6) = x^2 + x - 5 (x^4 - 12x^2 - x + 30) = (x + 2)(x - 3)(x^2 + x - 5) x^2 + x - 5 = 0 x^2 + x = 5 x^2 + x + 1/4= = 5 + 1/4 = 21/4 (x + 1/2)^2 = 21/4 x + 1/2 = ±√(21/4) x = -1/2 ± √(21/4) x = -2, 3, -1/2 + √(21/4), 1/2 - √(21/4)

[@SALogics]

Very nice! ❤

[@prollysine]

x^4+/-x^3-12x^2-x+30=0 , (x-3)(x^3+3x^2-3x-10)=0 , x=3 , x^3+3x^2-3x-10=0 , (x+2)(x^2+x-5)=0 , x=2 , 1 -3 1 2 x^2+x-5=0 , x= (-1+V(21))/2 , (-1-V(21))/2 , 3 -9 1 2 test , /not a solu x= -2 , (-1+V(21))/2/, -3 9 -5 -10 solu , x= 3 , (-1-V(21))/2 , -10 30

[@SALogics]

Very nice! ❤

[@labbhattacharjee108]

The solution by @SidneiMV in kzbin.info/www/bejne/q5XLq32FnZqkmpY is really smart

[@SALogics]

Very nice! ❤