If (x - 4)^2 = 81 and x > 0, what is the value of x? — SAT

If $(x - 4)^2 = 81$ and $x > 0$ , what is the value of $x$ ?

Answer: 13

The value of

x

is 13.

Taking the square root of both sides of

(x - 4)^2 = 81

yields

x - 4 = 9

x - 4 = -9

. Solving these linear equations gives

x = 13

x = -5

. Since

x > 0

is specified, the correct value of

x

is 13.

Take the square root of both sides of the equation.

x - 4 = 9

x - 4 = -9

Applying the square root property to solve the quadratic equation.

Solve each linear equation for

x

x = 13

x = -5

Adding 4 to both sides of each equation.

Apply the given constraint that

x > 0

x = 13

Since

-5

is not greater than 0, the only positive solution is 13.

Solving quadratic equations by taking square roots

Question