In the quadratic equation x^2 + kx + (k + 3) = 0, k is a constant. If the equation has exa… — SAT

In the quadratic equation $x^2 + kx + (k + 3) = 0$ , $k$ is a constant. If the equation has exactly one real solution, which of the following is a possible value of $k$ ?

A
-6
B
3
6Cevap
D
2

Cevap

The correct answer is 6. For the quadratic equation

x^2 + kx + (k + 3) = 0

to have exactly one real solution, its discriminant must be equal to zero. Setting the discriminant

b^2 - 4ac

to zero gives

k^2 - 4(1)(k + 3) = 0

, which simplifies to

k^2 - 4k - 12 = 0

. Factoring this equation yields

(k - 6)(k + 2) = 0

, meaning the possible values of

k

are

6

and

-2

. Among the choices, 6 is the only possible value listed.

Adım Adım Çözüm

Identify the coefficients of the given quadratic equation

x^2 + kx + (k + 3) = 0

The coefficients are

a = 1

b = k

, and

c = k + 3

To apply the discriminant formula, we need to identify the standard form coefficients

a

b

, and

c

Set the discriminant of the quadratic equation to zero.

The discriminant is

D = b^2 - 4ac

. Setting

D = 0

gives

k^2 - 4(1)(k + 3) = 0

, which simplifies to

k^2 - 4k - 12 = 0

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

Solve the quadratic equation

k^2 - 4k - 12 = 0

for

k

Factoring the quadratic yields

(k - 6)(k + 2) = 0

. Therefore, the possible values of

k

are

6

and

-2

Solving the equation reveals the values of the constant

k

that satisfy the condition.

Anahtar Kavram

Using the discriminant of a quadratic equation to determine the number of real solutions.

Alternatif Yöntem

Instead of factoring, the quadratic formula can be used to solve

k^2 - 4k - 12 = 0

, where

k = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm 8}{2}

, giving

k = 6

k = -2

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Konular

Bu konudan daha fazla soru