What are the solutions to the equation (1+x)^{12}=2.12 ?

The solutions to the equation (1+x)^{12}=2.12 are x=0.06462…,x=-2.06462…

Find the zeros of (1+x)^{12}=2.12

The zeros of (1+x)^{12}=2.12 are x=0.06462…,x=-2.06462…

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(1 + x)^{12} = 2.12

x = 0.06462 \dots, x = - 2.06462 \dots

Solution steps

(1 + x)^{12} = 2.12

For

(g (x))^{n} = f (a)

, n is even, the solutions are

Solve

The solutions are

x = 0.06462 \dots, x = - 2.06462 \dots

x^{3} = 10 - x

x, x^{4} + x^{2} y^{2} + y^{3} = 5

6 x^{5} - 12 x^{3} - 3 = 0

5 x^{3} + 8 x^{2} - 9 x + 2 = 0

x^{3} + 5 x^{2} + 3 x + 15 = 0

What are the solutions to the equation (1+x)^{12}=2.12 ?
The solutions to the equation (1+x)^{12}=2.12 are x=0.06462…,x=-2.06462…
Find the zeros of (1+x)^{12}=2.12
The zeros of (1+x)^{12}=2.12 are x=0.06462…,x=-2.06462…