What are the solutions to the equation (8x+1)^2=4 ?

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

Find the zeros of (8x+1)^2=4

The zeros of (8x+1)^2=4 are x= 1/8 ,x=-3/8

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(8 x + 1)^{2} = 4

x = \frac{1}{8}, x = - \frac{3}{8}

Decimal Notation

x = 0.125, x = - 0.375

Solution steps

(8 x + 1)^{2} = 4

For

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

the solutions are

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

Solve

8 x + 1 = 4 : x = \frac{1}{8}

Solve

8 x + 1 = - 4 : x = - \frac{3}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8}, x = - \frac{3}{8}

x^{2} - 2 = 2 - x^{2}

x^{2} - 2 a x + (a^{2} + b^{2}) = 0

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

63 - 16 x = - x^{2}

0 = 9 x^{2} - 3 x

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