MHB Solve $(2x+1)(3x+1)(5x+1)(30x+1)=10$: Real Solutions

  • Thread starter Thread starter anemone
  • Start date Start date
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find all real solutions of the equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$.
 
Mathematics news on Phys.org
anemone said:
Find all real solutions of the equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$.

we have $(2x+1)(30x+1)(3x+1)(5x+1) = 10$
or $(60x^2+ 32x + 1)(15x^2+ 8x + 1) = 10$

letting $15x^2 + 8x = t$

$(4t+1)(t+1) = 10$

or $4t^2 + 5 t + 1 = 10$

or $4t^2 + 5t - 9 = 0$

or $(4t+9)(t-1) =0 $



t = 1 or -9/4

t = 1 gives

$15x^2 + 8x-1=0$ giving $x = \dfrac{-4\pm\sqrt{31}}{15}$or $(15x^2 + 8x +\frac{9}4{4}) = 0$

or $(60x^2+ 32x + 9) = 0$

this gives complex solution

so solutions are $x = \dfrac{-4\pm\sqrt{31}}{15}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
6
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
3K
Replies
4
Views
4K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
8
Views
5K
Back
Top