If both the roots of the quadratic equation `x^(2)-2kx+k^(2)+k-5=0` are less than 5, then k lies in the interval.

Options

A. (5,6]
B. (6,`oo`)
C. `(-oo,4)`
D. `[4,5]`

Answer

To solve the problem, we need to find the values of \( k \) such that both roots of the quadratic equation

\[ x^2 - 2kx + (k^2 + k - 5) = 0 \]

are less than 5.

### Step-by-Step Solution:

**Step 1: Identify the quadratic equation.**

The given quadratic equation is

\[ x^2 - 2kx + (k^2 + k - 5) = 0. \]

**Step 2: Calculate the discriminant.**

The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by

\[ D = b^2 - 4ac. \]

Here, \( a = 1 \), \( b = -2k \), and \( c = k^2 + k - 5 \).

Calculating the discriminant:

\[ D = (-2k)^2 - 4 \cdot 1 \cdot (k^2 + k - 5) = 4k^2 - 4(k^2 + k - 5). \]

Expanding this gives:

\[ D = 4k^2 - 4k^2 - 4k + 20 = -4k + 20. \]

**Step 3: Ensure the discriminant is non-negative.**

For the roots to be real, the discriminant must be non-negative:

\[ -4k + 20 \geq 0. \]

Solving this inequality:

\[ 20 \geq 4k \implies k \leq 5. \]

**Step 4: Find the roots of the quadratic equation.**

Using the quadratic formula, the roots are given by:

\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2k \pm \sqrt{-4k + 20}}{2}. \]

This simplifies to:

\[ x = k \pm \sqrt{5 - k}. \]

**Step 5: Set conditions for the roots to be less than 5.**

For both roots to be less than 5, we require:

1. \( k + \sqrt{5 - k} < 5 \)
2. \( k - \sqrt{5 - k} < 5 \)

**Step 6: Solve the first inequality.**

From \( k + \sqrt{5 - k} < 5 \):

\[ \sqrt{5 - k} < 5 - k. \]

Squaring both sides (valid since both sides are non-negative):

\[ 5 - k < (5 - k)^2. \]

Expanding the right side:

\[ 5 - k < 25 - 10k + k^2. \]

Rearranging gives:

\[ 0 < k^2 - 9k + 20. \]

Factoring the quadratic:

\[ 0 < (k - 4)(k - 5). \]

This inequality holds for \( k < 4 \) or \( k > 5 \).

**Step 7: Solve the second inequality.**

From \( k - \sqrt{5 - k} < 5 \):

\[ -\sqrt{5 - k} < 5 - k \implies \sqrt{5 - k} > k - 5. \]

Squaring both sides:

\[ 5 - k > (k - 5)^2. \]

Expanding gives:

\[ 5 - k > k^2 - 10k + 25. \]

Rearranging gives:

\[ 0 > k^2 - 9k + 20. \]

Factoring:

\[ 0 > (k - 4)(k - 5). \]

This inequality holds for \( 4 < k < 5 \).

**Step 8: Combine the results.**

From the inequalities \( k < 4 \) and \( 4 < k < 5 \), we conclude that the valid interval for \( k \) is:

\[ k \in (-\infty, 4). \]

### Final Answer:

Thus, \( k \) lies in the interval \( (-\infty, 4) \).