If (2y - 1)^2 = 49 and y > 0, what is the value of y? — SAT

If $(2y - 1)^2 = 49$ and $y > 0$ , what is the value of $y$ ?

Answer: 4

The correct answer is

4

By taking the square root of both sides of the equation

(2y - 1)^2 = 49

, we get

2y - 1 = 7

2y - 1 = -7

. Solving

2y - 1 = 7

gives

2y = 8

, or

y = 4

. Solving

2y - 1 = -7

gives

2y = -6

, or

y = -3

. Since it is given that

y > 0

, the value of

y

must be

4

Take the square root of both sides of the equation.

2y - 1 = 7

2y - 1 = -7

Applying the square root property of equality to isolate the linear term.

Solve each linear equation for

y

y = 4

y = -3

Isolating

y

in each equation by adding

1

and then dividing by

2

Apply the constraint

y > 0

to identify the final value.

y = 4

The value

y = -3

is negative, so it does not satisfy the constraint

y > 0

Solving quadratic equations of the form

(ax + b)^2 = c

using the square root property.

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