Some interesting questions

(1)Solve:- α2+αβ+β2=γ2\alpha^2+\alpha\beta+\beta^2=\gamma^2 Where α,β,γ\alpha,\beta,\gamma are positive distinct primes

(2) Find out all roots of:- x4+9x3+1x^4+9x^3+1

(3) Find out the discriminant of:-

x3+x2+x+1x^3+x^2+x+1

(3) Solve the following differential equation:- ddx(y)=x3+x2y+xy2+y3x3+x2y+y3\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 xx such that:- x4+x3+x2+x+1x^4+x^3+x^2+x+1 is a perfect square

Note by Aman Sharma
4 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

3) ax3+bx2+cx+dax^3 + bx^2 + cx +d has discriminant as b2c24ac34b3d27a2d2+18abcdb^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd

Thus given discriminant is -16

U Z - 4 years, 9 months ago

Log in to reply

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

Pranjal Jain - 4 years, 9 months ago

Log in to reply

This Wiki might help you Thanks

U Z - 4 years, 9 months ago

Log in to reply

@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 - 4 years, 9 months ago

Log in to reply

@Pranjal Jain Need some time , involving @Deepanshu Gupta too

U Z - 4 years, 9 months ago

Log in to reply

@U Z Oh! Ok thanks!

Pranjal Jain - 4 years, 9 months ago

Log in to reply

dydx=1+yx+(yx)2+(yx)31+yx+(yx)3 \dfrac{dy}{dx} = \dfrac{1 + \dfrac{y}{x} + (\dfrac{y}{x})^2 + (\dfrac{y}{x})^3}{1 + \dfrac{y}{x} + (\dfrac{y}{x})^3}

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

xdvdx+v=1+v+v2+v3v3+v+1 x\dfrac{dv}{dx} + v = \dfrac{1 + v + v^2 + v^3}{v^3 + v + 1}

xdvdx=1v4v3+v+1 x\dfrac{dv}{dx} = \dfrac{1 - v^4}{v^3 + v + 1}

v3+v+11v4dv=1xdx \dfrac{v^3 + v + 1}{1 - v^4} dv = \dfrac{1}{x}dx

v(v2+1)(1+v2)(1v2)+1(1+v2)(1v2)=1xdx \dfrac{v(v^2+1)}{(1 + v^2)(1- v^2)} + \dfrac{1}{(1 + v^2)(1- v^2)} = \dfrac{1}{x}dx

v1+v2+12(11+v2+11v2)=1xdx \dfrac{v}{1 + v^2} + \dfrac{1}{2}( \dfrac{1}{1 + v^2} + \dfrac{1}{1 - v^2}) = \dfrac{1}{x}dx

=12ln(1+v2)+12(tan1v)+14(ln(1v1+v)=lnx+lnC \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 - 4 years, 9 months ago

Log in to reply

4) x5=1x^5 = 1

x4x5+x3x5+x2+x+1=0 \dfrac{x^4}{x^5} + \dfrac{x^3}{x^5} + x^2 + x + 1 =0

x2+1x2+x+1x+1=0 x^2 + \dfrac{1}{x^2} + x + \dfrac{1}{x} + 1 =0

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

U Z - 4 years, 9 months ago

Log in to reply

How did you get x5=1 x^5 = 1 ?

Siddhartha Srivastava - 4 years, 9 months ago

Log in to reply

You used the Fifth root of Unity here ? Right?

Where w5=1w^5 = 1

However, how will this give all possible values of x for which the given expression is a perfect square.

Sualeh Asif - 4 years, 9 months ago

Log in to reply

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

Siddhartha Srivastava - 4 years, 9 months ago

Log in to reply

@Siddhartha Srivastava That is what I asked @megh choksi. How does using x5=1x^5 = 1help 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 - 4 years, 9 months ago

Log in to reply

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

U Z - 4 years, 9 months ago

Log in to reply

You know this - sum of GP

U Z - 4 years, 9 months ago

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

Aman Sharma - 4 years, 9 months ago

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 α=2,Then β\alpha =2 , Then~ \beta will be an odd prime, and on RHS is also odd prime ,thus α=2 orβ=2\alpha=2~or\beta=2

4+2α+α2=γ2 4 + 2\alpha + \alpha^2 = \gamma^2

(α+1)2γ2=3 (\alpha + 1)^2 - \gamma^2 = -3

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

U Z - 4 years, 9 months ago

Log in to reply

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

Aman Sharma - 4 years, 9 months ago

Log in to reply

For 2, plot y=x4+9x3y=x^4+9x^3 and y=1y=-1 and check the points of intersection.

Pranjal Jain - 4 years, 9 months ago

Log in to reply

Very nice thinking

U Z - 4 years, 9 months ago

Log in to reply

Thanks!

Pranjal Jain - 4 years, 9 months ago

Log in to reply

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

Nik Arora - 4 years, 9 months ago

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.

Lu Chee Ket - 4 years, 9 months ago

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

Lu Chee Ket - 4 years, 9 months ago

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

Saurabh Patil - 4 years, 9 months ago

Log in to reply

Thanks , so the expression becomes xdvdx=1+v3v41+v3v4x\dfrac{dv}{dx} = \dfrac{1 + v^3 - v^4}{1 + v^3 - v^4}

v3+v+11+v3v4dv=1xdx \dfrac{v^3 + v + 1}{1 + v^3 - v^4} dv = \dfrac{1}{x}dx

1+v+v41+v3v4=1xdx 1 + \dfrac{v + v^4}{1 + v^3 - v^4} = \dfrac{1}{x}dx

then its simple

U Z - 4 years, 9 months ago

Log in to reply

Am I right? let α\alpha=a β\beta=b γ\gamma=c Then, a2+ab+b2c2a^2+ab+b^2-c^2=o a=b±b24b2+4c22\frac{-b \pm \sqrt{b^2-4b^2+4c^2}}{2}...........(1) The discriminant should be a perfect square.4c23b2\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 α\alpha , β\beta, γ\gamma.

naren ezhil - 4 years, 8 months ago

Log in to reply

discriminant -16??

Swaroop Muramalla - 4 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...