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# 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

Note by Aman Sharma
2 years, 2 months ago

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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 · 2 years, 2 months ago

Is there any general definition for "Determinant" for n-degree equation? · 2 years, 2 months ago

This Wiki might help you Thanks · 2 years, 2 months ago

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. · 2 years, 2 months ago

Need some time , involving @Deepanshu Gupta too · 2 years, 2 months ago

Oh! Ok thanks! · 2 years, 2 months ago

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

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

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

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

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

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

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

=$$\dfrac{1}{2}ln{(1+v^2)} + \dfrac{1}{2}(tan^{-1}v) + \dfrac{1}{4}(ln(\dfrac{1 - v}{1 + v}) = lnx + lnC$$ · 2 years, 2 months ago

4) $$x^5 = 1$$

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

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

$$Keep~ y = x + \dfrac{1}{x}$$ · 2 years, 2 months 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 · 2 years, 2 months ago

How did you get $$x^5 = 1$$? · 2 years, 2 months ago

You know this - sum of GP · 2 years, 2 months 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. · 2 years, 2 months ago

No where in the question does it mention that $$x$$ is the fifth root of unity. · 2 years, 2 months ago

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. · 2 years, 2 months ago

Sorry I wrote a method to find its roots · 2 years, 2 months 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 $$\alpha$$ , $$\beta$$, $$\gamma$$. · 2 years, 1 month 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 · 2 years, 2 months 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 · 2 years, 2 months 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 · 2 years, 2 months ago

For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident..... · 2 years, 2 months ago

You are right with 3, 5 and 7. I would like to answer to only question that I feel more interesting. · 2 years, 2 months ago

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

Very nice thinking · 2 years, 2 months ago

Thanks! · 2 years, 2 months ago

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

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

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

$$This ~can't~be ~possible~for~primes \to \boxed{(\alpha + \gamma + 1)(\alpha - \gamma + 1) = -3}$$ · 2 years, 2 months ago

Hint:-complete the square by adding $$\alpha\beta$$ both sides · 2 years, 2 months ago