If (x - 2)^2 - 9 = 0 and x < 0, what is the value of x^2 + 4x? — SAT

If $(x - 2)^2 - 9 = 0$ and $x < 0$ , what is the value of $x^2 + 4x$ ?

The value of the expression is -3

To find the value of

x^2 + 4x

, we first solve the equation

(x - 2)^2 - 9 = 0

for

x

. Adding 9 to both sides gives

(x - 2)^2 = 9

. Taking the square root of both sides yields

x - 2 = 3

x - 2 = -3

. Solving these two linear equations gives

x = 5

x = -1

. The problem states that

x < 0

, so we must choose

x = -1

. Substituting

x = -1

into the expression

x^2 + 4x

gives

(-1)^2 + 4(-1) = 1 - 4 = -3

Isolate the squared term in the equation.

(x - 2)^2 = 9

To solve a quadratic equation of the form

(x - h)^2 - k = 0

, we first add

k

to both sides to isolate the squared binomial.

Take the square root of both sides of the equation and solve for

x

x - 2 = \pm 3

, which gives

x = 5

x = -1

Taking the square root of both sides introduces both positive and negative roots, allowing us to find all possible values of

x

Apply the given constraint

x < 0

to determine the correct value of

x

x = -1

The problem specifies that

x

must be less than 0, so the positive solution

x = 5

must be rejected.

Substitute the value of

x

into the expression

x^2 + 4x

and simplify.

(-1)^2 + 4(-1) = 1 - 4 = -3

We substitute

x = -1

into the expression to find its final numerical value.

Solving quadratic equations by taking square roots and evaluating expressions under constraints

Estimated Time:1m 30s

Question