For what positive value of the constant k does the quadratic equation x^2 - 2kx + 3k + 10 … — GRE General Test

For what positive value of the constant $k$ does the quadratic equation $x^2 - 2kx + 3k + 10 = 0$ have exactly one real solution?

Answer: 5

The positive value of

k

is 5.

A quadratic equation has exactly one real solution when its discriminant

b^2 - 4ac

equals

0

. For

x^2 - 2kx + (3k + 10) = 0

, substituting

a = 1

b = -2k

, and

c = 3k + 10

gives

(-2k)^2 - 4(1)(3k + 10) = 0

, which simplifies to

4k^2 - 12k - 40 = 0

. Dividing the entire equation by

4

gives

k^2 - 3k - 10 = 0

. Factoring this quadratic yields

(k - 5)(k + 2) = 0

, giving solutions

k = 5

and

k = -2

. Since the question requires the positive value of

k

, the correct answer is

5

Identify the condition for a quadratic equation to have exactly one real solution.

The discriminant must equal zero (

b^2 - 4ac = 0

A quadratic equation

ax^2 + bx + c = 0

has a single repeated real root if and only if its discriminant is zero.

Set up the discriminant equation using the coefficients from

x^2 - 2kx + (3k + 10) = 0

Discriminant

= (-2k)^2 - 4(1)(3k + 10) = 4k^2 - 12k - 40 = 0

Here

a = 1

b = -2k

, and

c = 3k + 10

Divide by 4 and solve the quadratic in terms of

k

by factoring.

k^2 - 3k - 10 = 0 \Rightarrow (k - 5)(k + 2) = 0

, so

k = 5

k = -2

Factoring the simplified quadratic yields the two potential values of

k

Choose the value of

k

that satisfies the constraint in the stem.

k = 5

The question asks specifically for the positive value of

k

, so

k = -2

is rejected.

Quadratic Discriminant and Factoring

Estimated Time:1m 30s

Question