(1)Solve:- \[\alpha^2+\alpha\beta+\beta^2=\gamma^2\] Where \(\alpha,\beta,\gamma\) are positive distinct primes
(2) Find out all roots of:- \[x^4+9x^3+1\]
(3) Find out the discriminant of:-
\[x^3+x^2+x+1\]
(3) Solve the following differential equation:- \[\frac{d}{dx}(y)=\frac{x^3+x^2y+xy^2+y^3}{x^3+x^2y+y^3}\]
(4) Find out the sum of all values of \(x\) such that:- \(x^4+x^3+x^2+x+1\) is a perfect square
Note by
Aman Sharma
3 years, 8 months ago
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top Newest3) \(ax^3 + bx^2 + cx +d\) has discriminant as \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\)
Thus given discriminant is -16
Log in to reply
Is there any general definition for "Determinant" for n-degree equation?
Log in to reply
This Wiki might help you Thanks
Log in to reply
Is it possible to interpolate a function to calculate the number of terms in the discriminant? I tried to read wiki for this but couldn't understand it.
Log in to reply
Need some time , involving @Deepanshu Gupta too
Log in to reply
Oh! Ok thanks!
Log in to reply
\( \dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}\)
\( \dfrac{y}{x}= v\)
\( x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}\)
\( x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}\)
\( \dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx\)
\( \dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx\)
\( \dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx\)
=\( \dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC\)
Log in to reply
4) \(x^5 = 1\)
\( \dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0\)
\( x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0\)
\( Keep~ y = x + \dfrac{1}{x}\)
Log in to reply
But why you are finding root???? I mean how..will this give all values of x such that the given expression is perfect square.......i am not able to understand your solution..plz help
Log in to reply
How did you get \( x^5 = 1 \)?
Log in to reply
You know this - sum of GP
Log in to reply
You used the Fifth root of Unity here ? Right?
Where \(w^5 = 1\)
However, how will this give all possible values of x for which the given expression is a perfect square.
Log in to reply
No where in the question does it mention that \( x \) is the fifth root of unity.
Log in to reply
That is what I asked @megh choksi. How does using \(x^5 = 1\)help in getting pefect squared values for the expresion. In a hurry Megh must have misread the question, and found its roots what is not wanted at all.
Log in to reply
Sorry I wrote a method to find its roots
Log in to reply
Am I right? let \(\alpha\)=a \(\beta\)=b \(\gamma\)=c Then, \(a^2+ab+b^2-c^2\)=o a=\(\frac{-b \pm \sqrt{b^2-4b^2+4c^2}}{2}\)...........(1) The discriminant should be a perfect square.\(\Rightarrow 4c^2-3b^2\)=0 or a perfect square. \(\Rightarrow\) c=b= prime perfect square.(couldn't there be any other possibilities where b is not equal to c ?please check ) on substituting in (1) we can get values for \(\alpha\) , \(\beta\), \(\gamma\).
Log in to reply
Megh Choksi--- I think there is some mistake between the 3rd and 4th step!! as there is a term remaining in numerator that is v^3
Log in to reply
Thanks , so the expression becomes \(x\dfrac{dv}{dx} = \dfrac{1 + v^3 - v^4}{1 + v^3 - v^4}\)
\( \dfrac{v^3 + v + 1}{1 + v^3 - v^4} dv = \dfrac{1}{x}dx\)
\( 1 + \dfrac{v + v^4}{1 + v^3 - v^4} = \dfrac{1}{x}dx\)
then its simple
Log in to reply
2) For x^4 + 9 x^3 + 1 = 0 {Please write an equation.}
-0.489801491372372
-8.998627630183605
0.244214560777989 + j 0.408953683836186
0.244214560777989 - j 0.408953683836186
Log in to reply
For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident.....
Log in to reply
You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting.
Log in to reply
For 2, plot \(y=x^4+9x^3\) and \(y=-1\) and check the points of intersection.
Log in to reply
Very nice thinking
Log in to reply
Thanks!
Log in to reply
1) Please comment on this,
Rhs can't be even as sum of 3 odd primes will give always a prime,
If \(\alpha =2 , Then~ \beta\) will be an odd prime, and on RHS is also odd prime ,thus \(\alpha=2~or\beta=2\)
\( 4 + 2\alpha + \alpha^2 = \gamma^2\)
\( (\alpha + 1)^2 - \gamma^2 = -3\)
\( This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}\)
Log in to reply
Hint:-complete the square by adding \(\alpha\beta\) both sides
Log in to reply
discriminant -16??
Log in to reply