1.Let \(\alpha\) and \(\beta\) be the two roots of the cubic polynomial \(x^3+ax^2+bx+c\) satisfying \(\alpha\beta+1 = 0\).Prove that \(c^2+ac+b+1=0\) .

2.Solve for real \(x\): \(x^{\sqrt{x}}=\sqrt{x^x}\)

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

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

## Comments

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TopNewest\(1.\) Let the third root be \(\gamma\). \[\alpha\beta=-1\implies \alpha\beta\gamma=-\gamma.....(I)\] By Vieta's: \(\alpha\beta\gamma=-c\). Substituting in \(I\) to get \(\gamma=c\) and since \(\gamma\) is a root of cubic equation:- \[c^3+ac^2+bc+c=0\implies c^2+ac+b+1=0\]

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

Combining \(x=1,4\).

\(3.\) By Vieta's \(-D/A=(a-d)a(a+d)\)\(\implies \dfrac{-D}{aA}=(a-d)(a+d)=a^2-d^2\). \(\implies a^2+\dfrac{D}{aA}=a^2+(d^2-a^2)=d^2>0~~(d\neq 0)\). – Rishabh Cool · 11 months ago

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– Anik Mandal · 11 months ago

In the third question,there is a small typo.It should be \(a^2+\frac{D}{aA}\)Log in to reply

– Rishabh Cool · 11 months ago

Thanks .... corrected.Log in to reply

– Anik Mandal · 11 months ago

Thanks a lot for your reply!Log in to reply

What have you tried? – Calvin Lin Staff · 11 months ago

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– Anik Mandal · 11 months ago

I had tried using Vieta's and I have now got the solutions.Log in to reply