# Help needed!

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.

Note by Anik Mandal
2 years, 3 months ago

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$$1.$$ 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 $$-D/A=(a-d)a(a+d)$$$$\implies \dfrac{-D}{aA}=(a-d)(a+d)=a^2-d^2$$. $$\implies a^2+\dfrac{D}{aA}=a^2+(d^2-a^2)=d^2>0~~(d\neq 0)$$.

- 2 years, 3 months ago

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

- 2 years, 3 months ago

Thanks .... corrected.

- 2 years, 3 months ago

- 2 years, 3 months ago

What have you tried?

Staff - 2 years, 3 months ago

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

- 2 years, 3 months ago