(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
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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 – Megh Choksi · 2 years, 6 months ago
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Is there any general definition for "Determinant" for n-degree equation? – Pranjal Jain · 2 years, 6 months ago
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This Wiki might help you Thanks – Megh Choksi · 2 years, 6 months ago
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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. – Pranjal Jain · 2 years, 6 months ago
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Need some time , involving @Deepanshu Gupta too – Megh Choksi · 2 years, 6 months ago
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Oh! Ok thanks! – Pranjal Jain · 2 years, 6 months ago
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\( \dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}\)
\( \dfrac{y}{x}= v\)
\( x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}\)
\( x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}\)
\( \dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx\)
\( \dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx\)
\( \dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx\)
=\( \dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC\) – Megh Choksi · 2 years, 6 months ago
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4) \(x^5 = 1\)
\( \dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0\)
\( x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0\)
\( Keep~ y = x + \dfrac{1}{x}\) – Megh Choksi · 2 years, 6 months ago
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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 – Aman Sharma · 2 years, 6 months ago
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How did you get \( x^5 = 1 \)? – Siddhartha Srivastava · 2 years, 6 months ago
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You know this - sum of GP – Megh Choksi · 2 years, 6 months ago
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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. – Sualeh Asif · 2 years, 6 months ago
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No where in the question does it mention that \( x \) is the fifth root of unity. – Siddhartha Srivastava · 2 years, 6 months ago
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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. – Sualeh Asif · 2 years, 6 months ago
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Sorry I wrote a method to find its roots – Megh Choksi · 2 years, 6 months ago
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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 \(\alpha\) , \(\beta\), \(\gamma\). – Naren Ezhil · 2 years, 5 months ago
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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 – Saurabh Patil · 2 years, 6 months ago
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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 – Megh Choksi · 2 years, 6 months ago
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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 – Lu Chee Ket · 2 years, 6 months ago
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For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident..... – Nik Arora · 2 years, 6 months ago
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You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting. – Lu Chee Ket · 2 years, 6 months ago
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For 2, plot \(y=x^4+9x^3\) and \(y=-1\) and check the points of intersection. – Pranjal Jain · 2 years, 6 months ago
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Very nice thinking – Megh Choksi · 2 years, 6 months ago
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Thanks! – Pranjal Jain · 2 years, 6 months ago
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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 \(\alpha=2~or\beta=2\)
\( 4 + 2\alpha + \alpha^2 = \gamma^2\)
\( (\alpha + 1)^2 - \gamma^2 = -3\)
\( This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}\) – Megh Choksi · 2 years, 6 months ago
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Hint:-complete the square by adding \(\alpha\beta\) both sides – Aman Sharma · 2 years, 6 months ago
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discriminant -16?? – Swaroop Muramalla · 2 years, 6 months ago
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