What are the solutions to the equation (4x-1)^2=27 ?

The solutions to the equation (4x-1)^2=27 are x=(3sqrt(3)+1)/4 ,x=(-3sqrt(3)+1)/4

Find the zeros of (4x-1)^2=27

The zeros of (4x-1)^2=27 are x=(3sqrt(3)+1)/4 ,x=(-3sqrt(3)+1)/4

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(4 x - 1)^{2} = 27

x = \frac{3 3 + 1}{4}, x = \frac{- 3 3 + 1}{4}

Decimal

x = 1.54903 \dots, x = - 1.04903 \dots

Solution steps

(4 x - 1)^{2} = 27

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

4 x - 1 = 27 : x = \frac{3 3 + 1}{4}

Solve

4 x - 1 = - 27 : x = \frac{- 3 3 + 1}{4}

The solutions to the quadratic equation are:

x = \frac{3 3 + 1}{4}, x = \frac{- 3 3 + 1}{4}

k^{2} + 12 (- 2) = k (- 2)

120 = x^{2} - 1

x, x^{2} - y^{2} + 4 x + 4 = 0

(5 - 2 x)^{2} - 16 = 0

3 b^{2} - 4 b + 1 = 0

What are the solutions to the equation (4x-1)^2=27 ?
The solutions to the equation (4x-1)^2=27 are x=(3sqrt(3)+1)/4 ,x=(-3sqrt(3)+1)/4
Find the zeros of (4x-1)^2=27
The zeros of (4x-1)^2=27 are x=(3sqrt(3)+1)/4 ,x=(-3sqrt(3)+1)/4