What are the solutions to the equation (3x^2+5x-7)^4=1 ?

The solutions to the equation (3x^2+5x-7)^4=1 are x=1,x=-8/3 ,x=(-5+sqrt(97))/6 ,x=(-5-sqrt(97))/6

Find the zeros of (3x^2+5x-7)^4=1

The zeros of (3x^2+5x-7)^4=1 are x=1,x=-8/3 ,x=(-5+sqrt(97))/6 ,x=(-5-sqrt(97))/6

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(3 x^{2} + 5 x - 7)^{4} = 1

x = 1, x = - \frac{8}{3}, x = \frac{- 5 + 97}{6}, x = \frac{- 5 - 97}{6}

Decimal Notation

x = 1, x = - 2.66666 \dots, x = 0.80814 \dots, x = - 2.47480 \dots

Solution steps

(3 x^{2} + 5 x - 7)^{4} = 1

For

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

, n is even, the solutions are

Solve

The solutions are

x = 1, x = - \frac{8}{3}, x = \frac{- 5 + 97}{6}, x = \frac{- 5 - 97}{6}

6 x^{4} - 35 x^{3} + 62 x^{2} - 35 x + 6 = 0

21 x^{3} + x^{2} - 5 x - 1 = 0

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

0 = x^{4} - 10 x^{2} + 9

- 3 + c^{3} = 73

What are the solutions to the equation (3x^2+5x-7)^4=1 ?
The solutions to the equation (3x^2+5x-7)^4=1 are x=1,x=-8/3 ,x=(-5+sqrt(97))/6 ,x=(-5-sqrt(97))/6
Find the zeros of (3x^2+5x-7)^4=1
The zeros of (3x^2+5x-7)^4=1 are x=1,x=-8/3 ,x=(-5+sqrt(97))/6 ,x=(-5-sqrt(97))/6