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Q-141663-by-ajfour-sir-reposted-x-2-x-12-x-15-k-x-16-k-gt-0-Find-x-in-terms-of-k-

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Question Number 142939 by Rasheed.Sindhi last updated on 07/Jun/21
Q#141663 by ajfour sir reposted. x^2 (x−12)(x−15)=k(x−16) ;k>0 Find x in terms of k.
You can't use 'macro parameter character #' in math modex2(x−12)(x−15)=k(x−16);k>0Findxintermsofk.
Answered by Rasheed.Sindhi last updated on 07/Jun/21
{x^2 (x−12)(x−15)}/(x−16)=k Quotient is k with no remainder. Let p(x)=x^2 (x^2 −27x+180) =x^4 −27x^3 +180x^2 +0x+0 By synthetic division: determinant (((16)),1,(−27),(180),0,0),(,,( 16),(−176),(64),(1024)),(,1,(−11),( 4),(64),(1024))) p(x)=Q(x)D(x)+R x^2 (x−12)(x−15)= (x^3 −11x^2 +4x+64)(x−16)+1024 Neither the quotient is constant nor the remainder is zero ∀ x So this proves that the equation has no solution.
{x2(x−12)(x−15)}/(x−16)=kQuotientiskwithnoremainder.Letp(x)=x2(x2−27x+180)=x4−27x3+180x2+0x+0Bysyntheticdivision:16)1−271800016−1766410241−114641024p(x)=Q(x)D(x)+Rx2(x−12)(x−15)=(x3−11x2+4x+64)(x−16)+1024Neitherthequotientisconstantnortheremainderiszero∀xSothisprovesthattheequationhasnosolution.
Commented by mr W last updated on 07/Jun/21
i think it′s wrong somewhere sir. y=x^2 (x−12)(x−15) is a curve passing through (0,0),(12,0) and (15,0) y=k(x−16) is a line passing through (16,0) with inclination k. from (16,0) we can draw two tangent lines to the curve, say these tangent lines have inclination k_1 and k_2 . we know for k≤k_1 or k≥k_2 , the line y=k(x−16) intersects the curve at least at one point and at most at four points. that means for k≤k_1 or k≥k_2 , the equation x^2 (x−12)(x−15)=k(x−16) has at least one solution and at most four solutions.
ithinkit′swrongsomewheresir.y=x2(x−12)(x−15)isacurvepassingthrough(0,0),(12,0)and(15,0)y=k(x−16)isalinepassingthrough(16,0)withinclinationk.from(16,0)wecandrawtwotangentlinestothecurve,saythesetangentlineshaveinclinationk1andk2.weknowfork⩽k1ork⩾k2,theliney=k(x−16)intersectsthecurveatleastatonepointandatmostatfourpoints.thatmeansfork⩽k1ork⩾k2,theequationx2(x−12)(x−15)=k(x−16)hasatleastonesolutionandatmostfoursolutions.
Commented by mr W last updated on 07/Jun/21
Commented by Rasheed.Sindhi last updated on 08/Jun/21
ThanX for your Efforts Sir! Perhaps my logic is only valid for integer k(?).
ThanXforyourEffortsSir!Perhapsmylogicisonlyvalidforintegerk(?).
Commented by mr W last updated on 08/Jun/21
the question doesn′t request if x^2 (x−12)(x−15) can be expressed as k(x−16) for any values of x, but only asks if one or more values of x exist such that x^2 (x−12)(x−15)=k(x−16).
thequestiondoesn′trequestifx2(x−12)(x−15)canbeexpressedask(x−16)foranyvaluesofx,butonlyasksifoneormorevaluesofxexistsuchthatx2(x−12)(x−15)=k(x−16).
Commented by Rasheed.Sindhi last updated on 08/Jun/21
OK Sir, ThanX again.
OKSir,ThanXagain.
Answered by Rasheed.Sindhi last updated on 07/Jun/21
x^2 (x−12)(x−15)=k(x−16) ;k>0 This means : When x^2 (x−12)(x−15) is divided by x−16, The quotient is k(>0) and the remainder is 0 But we can verify that the remainder R is: R=(16)^2 (16−12)(16−15) =256.4.1=1024≠0 So for any value of x the remainder can′t be zero. This proves that the above equation can′t be solved (even in terms of k>0.)
x2(x−12)(x−15)=k(x−16);k>0Thismeans:Whenx2(x−12)(x−15)isdividedbyx−16,Thequotientisk(>0)andtheremainderis0ButwecanverifythattheremainderRis:R=(16)2(16−12)(16−15)=256.4.1=1024≠0Soforanyvalueofxtheremaindercan′tbezero.Thisprovesthattheaboveequationcan′tbesolved(evenintermsofk>0.)

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