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

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The equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$ is analyzed for real solutions. By expanding and simplifying the left side, it is transformed into a polynomial equation. Various methods, including numerical approximations and graphical analysis, are suggested to find the roots. The discussion emphasizes the importance of checking for rational solutions and using tools like the Intermediate Value Theorem. Ultimately, the focus remains on identifying all real solutions to the equation.
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Find all real solutions of the equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$.
 
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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 $x = \dfrac{-4\pm\sqrt{31}}{15}$
 
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