In the quadratic equation x^2 - kx + 36 = 0, k is a positive constant. If the equation has… — SAT

In the quadratic equation $x^2 - kx + 36 = 0$ , $k$ is a positive constant. If the equation has exactly one real solution, what is the value of $k$ ?

Answer: 12

The value of

k

12

The quadratic equation

x^2 - kx + 36 = 0

has exactly one real solution when its discriminant is equal to zero. The discriminant of

ax^2 + bx + c = 0

b^2 - 4ac

. For this equation,

a = 1

b = -k

, and

c = 36

. Setting the discriminant to zero gives

(-k)^2 - 4(1)(36) = 0

, which simplifies to

k^2 - 144 = 0

. Solving for

k

gives

k^2 = 144

, so

k = 12

k = -12

. Since the problem states that

k

is a positive constant, the correct value of

k

12

Identify the coefficients of the quadratic equation

x^2 - kx + 36 = 0

a = 1

b = -k

, and

c = 36

To apply the discriminant formula, we need the values of

a

b

, and

c

Set the discriminant

\Delta = b^2 - 4ac

0

(-k)^2 - 4(1)(36) = 0

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

Solve the equation for

k

k^2 - 144 = 0 \implies k^2 = 144 \implies k = 12

(since

k

is a positive constant).

Solving the equation yields

k = \pm 12

, and we select the positive value as specified in the problem.

Discriminant of a quadratic equation

Question