I attempted a question which says:

\(f(x)=3x^3 +2x^2 +14x -5.\)

a = sum of real roots. Find 93a. I guessed the answer to be 31 and it was correct. Can anyone help me to solve it actually?

I have one more question. Can't we view the solutions of the practice problems we are unable to solve?

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## Comments

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TopNewestNotice that: \[3x^3 + 2x^2 + 14x -5 \equiv (3x-1)(x^2+x+5)\] Consider the nature of the roots of \(x^2 + x + 5\). We will use the discriminant for this: \[1^2- 4 \times 1 \times 5 = -19 < 0\] Therefore, it is clear the \(x^2 + x + 5\) has no real roots. From this we can see that the only real root is when: \[3x - 1 = 0 \Rightarrow x = \frac{1}{3}\] Therefore: \[a = \frac{1}{3} \Rightarrow 93a = \fbox{31}\]

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Thanks!

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I can't help with the question but I am curious as to how did you even guess that?

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Maharnab must have guessed it as question was positive integer type. 93a is asked means a might not be integer but 93a must be. So factors of 93 are limited...

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Is this a question included in the Practice section?

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Yes, it is. Is its solution available at Brilliant.org?

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On the page for the question on the top right there is a grey icon like two chain links connected, like the on the bottom right of the original post. If you put the mouse on it it says "copy URL". Click copy and paste it into the website search bar. It should go to another page with the same problem and a button with reveal solution on it.

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Just to say, every time I did this it worked but a glitch or something stops you from doing that problem set again. If you click "Next Problem" it doesn't work. I hope they fix it soon.

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Can you give the link to it?

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We can see here that \(\frac{1}{3}\) is a zero of the polynomial.Then we can divide the polynomial by \(3x -1\) and get the quadratic polynomial \(x^{2}+x+5\), which is having all its zeroes imaginary.Hence, we get the result.

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If certain fractions are the zeros of a polynomial, is there a better way to find them other than by observing and performing trials?

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since the cubic polynomial may have 3 real roots or 1 real root but never 2 real roots as in that case the rest complex root will not have any conjugate pair left. now for three real roots we get a= (-2/3) i.e

93a= -62 which is impossible as the answer must be a positive integer.[just for checking] and then being convinced with the fact that only one real root is possible i used the factorisation method same as OLIVER W. and finally got the answer 93a= 31

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graphically we can prove that a=1/3

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Isn't there a cubic formula to get the roots of a cube without trial and error?

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I'm just curious is there any formula or something that can directly compute the value of x in this like the quadratic equation ?

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