Some interesting questions

(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

#Algebra #NumberTheory #Calculus #ProblemSolving #DifferentialEquations

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Comments

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3) $ax^3 + bx^2 + cx +d$ has discriminant as $b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$

Thus given discriminant is -16

U Z - 5 years ago

Is there any general definition for "Determinant" for n-degree equation?

Pranjal Jain - 5 years ago

This Wiki might help you Thanks

U Z - 5 years ago

@U Z – 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 - 5 years ago

$\dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}$

$\dfrac{y}{x}= v$

$x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}$

$x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}$

$\dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx$

$\dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx$

$\dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx$

= $\dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC$

U Z - 5 years ago

4) $x^5 = 1$

$\dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0$

$x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0$

$Keep~ y = x + \dfrac{1}{x}$

U Z - 5 years ago

How did you get $x^5 = 1$ ?

Siddhartha Srivastava - 5 years ago

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 - 5 years ago

@Sualeh Asif – No where in the question does it mention that $x$ is the fifth root of unity.

Siddhartha Srivastava - 5 years ago

@Siddhartha Srivastava – 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 - 5 years ago

@Sualeh Asif – Sorry I wrote a method to find its roots

U Z - 5 years ago

You know this - sum of GP

U Z - 5 years ago

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 - 5 years ago

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 $\alpha=2~or\beta=2$

$4 + 2\alpha + \alpha^2 = \gamma^2$

$(\alpha + 1)^2 - \gamma^2 = -3$

$This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}$

U Z - 5 years ago

Hint:-complete the square by adding $\alpha\beta$ both sides

Aman Sharma - 5 years ago

For 2, plot $y=x^4+9x^3$ and $y=-1$ and check the points of intersection.

Pranjal Jain - 5 years ago

Very nice thinking

U Z - 5 years ago

Thanks!

Pranjal Jain - 5 years ago

For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident.....

Nik Arora - 5 years ago

You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting.

Lu Chee Ket - 5 years ago

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 - 5 years ago

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 - 5 years ago

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

U Z - 5 years ago

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 - 4 years, 12 months ago

discriminant -16??

Swaroop Muramalla - 5 years 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}`	$\boxed{123}$