National Camp Exam 2018 P9

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Swapnil Barua
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Joined:Tue Apr 13, 2021 2:55 pm
National Camp Exam 2018 P9

Unread post by Swapnil Barua » Tue Apr 20, 2021 8:57 pm

The equation \begin{align*}
9x³-3x²-3x-1 & =0
\end{align*} has a real root of the form $\frac{√3a+√3b+1}{c}$
where
a, b, c are positive integers. Find a + b + c.

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: National Camp Exam 2018 P9

Unread post by Mehrab4226 » Wed Apr 21, 2021 12:28 pm

Swapnil Barua wrote:
Tue Apr 20, 2021 8:57 pm
The equation \begin{align*}
9x³-3x²-3x-1 & =0
\end{align*} has a real root of the form $\frac{√3a+√3b+1}{c}$
where
a, b, c are positive integers. Find a + b + c.
I think you didn't type the question correctly.
Short solution:
Given,
$9x^3-3x^2-3x-1=0$
Or,$10x^3-x^3-3x^2-3x-1=0$
Or,$10x^3-(x+1)^3=0 \cdots (1)$
Or,$10x^3=(x+1)^3$
Or,$x \sqrt[3]{10}=x+1$
Or,$x \sqrt[3]{10}-x=1$
Or,$x(\sqrt[3]{10}-1)=1$
Or,$x=\frac{1}{(\sqrt[3]{10}-1)}$
Or,$x=\frac{(\sqrt[3]{100}+\sqrt[3]{10}+1)}{(\sqrt[3]{10}-1)((\sqrt[3]{100}+\sqrt[3]{10}+1)}$
Or,$x=\frac{\sqrt[3]{100}+\sqrt[3]{10}+1}{10-1}$
$\therefore a=100,b=10, c=9$
$\therefore 1+b+c=100+10+9=119$
Long solution for which the question isn't typed correctly:
$10x^3-(x+1)^3=0 \cdots (1)$
Or,$(x\sqrt[3]{10})^3-(x+1)^3=0$
Or,$(x\sqrt[3]{10}-x-1)(x^2\sqrt[3]{100}+x^2\sqrt[3]{10}+x\sqrt[3]{10}+x+1)=0$
Either,
$(x\sqrt[3]{10}-x-1)=0$
This will lead to the soluton above
Or,
$(x^2\sqrt[3]{100}+x^2\sqrt[3]{10}+x\sqrt[3]{10}+x+1)=0$
$x^2(\sqrt[3]{100}+\sqrt[3]{10})+x(\sqrt[3]{10}+1)+1=0$
Now the discriminant is,
$(\sqrt[3]{10}+1)2-4(\sqrt[3]{100}+\sqrt[3]{10})$
$=\sqrt[3]{100}+2\sqrt[3]{10}+1-4\sqrt[3]{100}-4\sqrt[3]{10}$
$=1-3\sqrt[3]{100}-2\sqrt[3]{10}<0$
So there are no other real roots.

The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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