What are the solutions to the equation (2x+1)^2=81 ?

The solutions to the equation (2x+1)^2=81 are x=4,x=-5

Find the zeros of (2x+1)^2=81

The zeros of (2x+1)^2=81 are x=4,x=-5

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(2 x + 1)^{2} = 81

x = 4, x = - 5

Solution steps

(2 x + 1)^{2} = 81

For

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

the solutions are

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

Solve

2 x + 1 = 81 : x = 4

Solve

2 x + 1 = - 81 : x = - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

(2 x - 4)^{2} - 10 (2 x - 4) + 21 = 0

2 x^{2} = 3 x

\frac{16}{3} + \frac{16}{15} + \frac{16}{35}

\frac{6}{\frac{3}{4}}

\frac{1}{4} \cdot 2

What are the solutions to the equation (2x+1)^2=81 ?
The solutions to the equation (2x+1)^2=81 are x=4,x=-5
Find the zeros of (2x+1)^2=81
The zeros of (2x+1)^2=81 are x=4,x=-5