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1.Let α\alpha and β\beta be the two roots of the cubic polynomial x3+ax2+bx+cx^3+ax^2+bx+c satisfying αβ+1=0\alpha\beta+1 = 0.Prove that c2+ac+b+1=0c^2+ac+b+1=0 .

2.Solve for real xx: xx=xxx^{\sqrt{x}}=\sqrt{x^x}

3.If polynomial p(x)=Ax3+Bx2+Cx+Dp(x) = Ax^3+Bx^2+Cx+D vanishes at x=ad,a,a+dx=a-d,a,a+d then prove that a2+DaA>0a^2+\frac{D}{aA}>0.Here A,B,C,DA,B,C ,D are constants.

Please provide a solution to these problems as soon as possible.Thanks.

Note by Anik Mandal
3 years, 5 months ago

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1.1. Let the third root be γ\gamma. αβ=1    αβγ=γ.....(I)\alpha\beta=-1\implies \alpha\beta\gamma=-\gamma.....(I) By Vieta's: αβγ=c\alpha\beta\gamma=-c. Substituting in II to get γ=c\gamma=c and since γ\gamma is a root of cubic equation:- c3+ac2+bc+c=0    c2+ac+b+1=0c^3+ac^2+bc+c=0\implies c^2+ac+b+1=0


2.2. We always look for 0,10,1 in these equations. Obviously x=1x=1 is a solution. Now, xx=xx/2\Large{x^{\sqrt x}=x^{x/2}}     x=x/2\Large\implies \sqrt x=x/2     x(x2)=0\Large \implies \sqrt x(\sqrt x-2)=0     x=2    x=4\Large \implies \sqrt x=2\implies x=4 (x=0x=0 is obviously rejected due to indeterminate form)

Combining x=1,4x=1,4.


3.3. By Vieta's D/A=(ad)a(a+d)-D/A=(a-d)a(a+d)    DaA=(ad)(a+d)=a2d2\implies \dfrac{-D}{aA}=(a-d)(a+d)=a^2-d^2.     a2+DaA=a2+(d2a2)=d2>0  (d0)\implies a^2+\dfrac{D}{aA}=a^2+(d^2-a^2)=d^2>0~~(d\neq 0).

Rishabh Jain - 3 years, 5 months ago

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In the third question,there is a small typo.It should be a2+DaAa^2+\frac{D}{aA}

Anik Mandal - 3 years, 5 months ago

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Thanks .... corrected.

Rishabh Jain - 3 years, 5 months ago

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Thanks a lot for your reply!

Anik Mandal - 3 years, 5 months ago

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What have you tried?

Calvin Lin Staff - 3 years, 5 months ago

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I had tried using Vieta's and I have now got the solutions.

Anik Mandal - 3 years, 5 months ago

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