(1)Solve:-
α2+αβ+β2=γ2
Where α,β,γ are positive distinct primes
(2)
Find out all roots of:-
x4+9x3+1
(3)
Find out the discriminant of:-
x3+x2+x+1
(3)
Solve the following differential equation:-
dxd(y)=x3+x2y+y3x3+x2y+xy2+y3
(4)
Find out the sum of all values of x such that:-
x4+x3+x2+x+1 is a perfect square
Easy Math Editor
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Top Newest3) ax3+bx2+cx+d has discriminant as b2c2−4ac3−4b3d−27a2d2+18abcd
Thus given discriminant is -16
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Is there any general definition for "Determinant" for n-degree equation?
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This Wiki might help you Thanks
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@Deepanshu Gupta too
Need some time , involvingLog in to reply
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dxdy=1+xy+(xy)31+xy+(xy)2+(xy)3
xy=v
xdxdv+v=v3+v+11+v+v2+v3
xdxdv=v3+v+11−v4
1−v4v3+v+1dv=x1dx
(1+v2)(1−v2)v(v2+1)+(1+v2)(1−v2)1=x1dx
1+v2v+21(1+v21+1−v21)=x1dx
=21ln(1+v2)+21(tan−1v)+41(ln(1+v1−v)=lnx+lnC
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4) x5=1
x5x4+x5x3+x2+x+1=0
x2+x21+x+x1+1=0
Keep y=x+x1
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How did you get x5=1?
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You used the Fifth root of Unity here ? Right?
Where w5=1
However, how will this give all possible values of x for which the given expression is a perfect square.
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x is the fifth root of unity.
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x5=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.
That is what I asked @megh choksi. How does usingLog in to reply
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You know this - sum of GP
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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
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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 α=2,Then β will be an odd prime, and on RHS is also odd prime ,thus α=2 orβ=2
4+2α+α2=γ2
(α+1)2−γ2=−3
This can′t be possible for primes→(α+γ+1)(α−γ+1)=−3
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Hint:-complete the square by adding αβ both sides
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For 2, plot y=x4+9x3 and y=−1 and check the points of intersection.
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Very nice thinking
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Thanks!
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For (1).. 3,5,7 satisfies the condition.... @megh choksi sum of odd primes is not always prime as evident.....
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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.
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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
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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
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Thanks , so the expression becomes xdxdv=1+v3−v41+v3−v4
1+v3−v4v3+v+1dv=x1dx
1+1+v3−v4v+v4=x1dx
then its simple
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Am I right? let α=a β=b γ=c Then, a2+ab+b2−c2=o a=2−b±b2−4b2+4c2...........(1) The discriminant should be a perfect square.⇒4c2−3b2=0 or a perfect square. ⇒ 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 , β, γ.
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discriminant -16??
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