What are the solutions to the equation (2x+1)^2=121 ?

The solutions to the equation (2x+1)^2=121 are x=5,x=-6

Find the zeros of (2x+1)^2=121

The zeros of (2x+1)^2=121 are x=5,x=-6

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(2 x + 1)^{2} = 121

x = 5, x = - 6

Solution steps

(2 x + 1)^{2} = 121

For

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

the solutions are

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

Solve

2 x + 1 = 121 : x = 5

Solve

2 x + 1 = - 121 : x = - 6

The solutions to the quadratic equation are:

x = 5, x = - 6

0 = 2 x^{2} - 4

3 x^{2} + 25 - 7 x = 2 x^{2} + 15

x (5 x - 6) = - 1

2 x (3 x - 1) = 8

k^{2} - 7 k - 8 = 0

What are the solutions to the equation (2x+1)^2=121 ?
The solutions to the equation (2x+1)^2=121 are x=5,x=-6
Find the zeros of (2x+1)^2=121
The zeros of (2x+1)^2=121 are x=5,x=-6