The Radical Equation That STUMPS 99% of Students!

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  • Опубліковано 17 лис 2024

КОМЕНТАРІ • 10

  • @RashmiRay-c1y
    @RashmiRay-c1y Місяць тому +1

    x is positive and less than 2. After squaring, the equation can be written as x^4-18x^3+72x^2+81x-162=0. Noting that -162 = (9)(-18), we write the quartic equation as (x^2+a-18)(x^2+b+9)=0 and by comparison, a -9, b= -9. So (x^2-9x-18)(x^2-9x+9)=0 > x= 1/2[9+/-3√17], x=1/2[9+/-3√5]. But x is positive and less than 2 > x=1/2[9-3√5].

  • @tunneloflight
    @tunneloflight Місяць тому

    Oh, by the way. In real world control systems where equations like this are governing, the systems may be intended for example to operate at the first solution value with x < 2. But when system perturbations happen two sets of results occur. the first state is a system that orbits the first result. With a larger upset or perturbation the values deviate and the system may return to stability, or it may seek out either or both of the other solutions.
    It can settle at either, or if there are sufficient perturbations it can result in a quasi bistable state where the system oscillates between the two solutions, or orbits them.
    Imagine if this occurred in a particle board plant. The result could be finished product where some parameter - density, thickness, toughness, elasticity oscillates around these three values with little explanation and creating difficulty for the operators, with lots of defective or non-specification products. Engineers, operators and management would all be pulling there hair out trying to figure it out.
    Or, the nominator and denominator on the left side may represent physical components that simply cannot exist in the negative condition. If they do, parts, levers, linkages, ... or other things may break with terrible consequences as the equation then runs wild. I've seen that happen. And if it is in an obscure place, the breakage may not be apparent, other than through the process running wild.
    Or, for example, the imaginary roots might be the equivalent of two parts working in opposition like elbows and knees, that should only ever work in the forward-backward plane - instead being driven to turn at right angles to work in the left-right plane, though in driving elevation the result is the same.

  • @bobross7473
    @bobross7473 Місяць тому

    Oh my god this was insane but I nailed it! I’ve done a ton of practice with these kinds of problems so it paid off. I didn’t immediately get the common (x squared - 9x) term but after some messing around I managed to get it. Beautiful challenge, but hard indeed.

  • @АндрейПергаев-з4н
    @АндрейПергаев-з4н Місяць тому

    Если 0

  • @Shobhamaths
    @Shobhamaths Місяць тому

    On squaring both sides
    x^4-18x^3+72x^2+18x-162=0 if we simplify by making two quadratic equations
    x=(9±3√5)/2,x>0
    x=(9+3√5)2👍🏻

  • @tunneloflight
    @tunneloflight Місяць тому

    Actually all three solutions are valid. The solution set is bifurcated with 0

  • @Quest3669
    @Quest3669 Місяць тому +1

    x=(9-√45)/2= 1.146 only soln

  • @anestismoutafidis4575
    @anestismoutafidis4575 Місяць тому

    2-x/(x-6)•(x-12)-x^2/81=0
    2- |-0,9|/(|-0,9| -6)•(|-0,9|-12) =
    (1,1)^1/2 ÷ (89,01)^1/2 = 1,04/9,43
    =1/9 x=|-0,9|

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 Місяць тому

    (2)^2 ➖ (x)^2/(x ➖ 6)^2(x ➖ 12)^2= {4 ➖ x^2}/(x^2 ➖ 36)(x^2 ➖ 144)=2/{34+142}/2/176 2/2^88 1/1^2^44 /2^2^22 /1^1^2^11 /2^11^1/ 2^1^1 2^1(x ➖ 2x+1){x+x ➖}/{9x9x ➖}=x^2/18x^2 x^1/3^6x^1 /3^3^2x^1 1^3^2x^1 3^2x (x ➖ 3x+2).