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.

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

$1.$ 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 Jain - 5 years, 1 month ago

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

Anik Mandal - 5 years, 1 month ago

Thanks .... corrected.

Rishabh Jain - 5 years, 1 month ago

Thanks a lot for your reply!

Anik Mandal - 5 years, 1 month ago

What have you tried?

Calvin Lin Staff - 5 years, 1 month ago

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

Anik Mandal - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$