What are the solutions to the equation 2(4x)^2=128 ?

The solutions to the equation 2(4x)^2=128 are x=2,x=-2

Find the zeros of 2(4x)^2=128

The zeros of 2(4x)^2=128 are x=2,x=-2

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2 (4 x)^{2} = 128

x = 2, x = - 2

Solution steps

2 (4 x)^{2} = 128

Divide both sides by

2

(4 x)^{2} = 64

For

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

the solutions are

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

Solve

4 x = 64 : x = 2

Solve

4 x = - 64 : x = - 2

The solutions to the quadratic equation are:

x = 2, x = - 2

3 p^{2} + 5 p - 10 = 0

4 x = 21 - x^{2}

2 n^{2} - 2 n + 7 = 0

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

4 x^{2} - 32 x + 65 = 0

What are the solutions to the equation 2(4x)^2=128 ?
The solutions to the equation 2(4x)^2=128 are x=2,x=-2
Find the zeros of 2(4x)^2=128
The zeros of 2(4x)^2=128 are x=2,x=-2