What are the solutions to the equation 35=(1+x)^4 ?

The solutions to the equation 35=(1+x)^4 are x=\sqrt[4]{35}-1,x=-\sqrt[4]{35}-1

Find the zeros of 35=(1+x)^4

The zeros of 35=(1+x)^4 are x=\sqrt[4]{35}-1,x=-\sqrt[4]{35}-1

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35 = (1 + x)^{4}

Decimal Notation

x = 1.43229 \dots, x = - 3.43229 \dots

Solution steps

35 = (1 + x)^{4}

Switch sides

(1 + x)^{4} = 35

For

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

, n is even, the solutions are

Solve

The solutions are

234514 x^{3} - 746760 x + 395221 = 0

(r - 1)^{4} (r + 1) = 0

x^{4} + x^{3} = 3 - 4 x

2 x^{3} - 3 x + 6 = 0

2 x^{3} + 6 x^{2} + 6 x = 0

What are the solutions to the equation 35=(1+x)^4 ?
The solutions to the equation 35=(1+x)^4 are x=\sqrt[4]{35}-1,x=-\sqrt[4]{35}-1
Find the zeros of 35=(1+x)^4
The zeros of 35=(1+x)^4 are x=\sqrt[4]{35}-1,x=-\sqrt[4]{35}-1