Context: Let ‘k’ be an integer such that one of the roots of the quadratic equation 4x2 – 4(√3 + 1)x + √3k = 24 is an integer. What is the sum of the values of ‘k’?

Answer question

We are given the quadratic equation:

\[
4x^2 - 4(\sqrt{3} + 1)x + \sqrt{3}k = 24
\]

First, we can rewrite the equation in the standard form \( ax^2 + bx + c = 0 \) by moving the constant term to the left-hand side:

\[
4x^2 - 4(\sqrt{3} + 1)x + (\sqrt{3}k - 24) = 0
\]

The equation now is:

\[
4x^2 - 4(\sqrt{3} + 1)x + (\sqrt{3}k - 24) = 0
\]

Let's divide the entire equation by 4 to simplify it:

\[
x^2 - (\sqrt{3} + 1)x + \frac{\sqrt{3}k - 24}{4} = 0
\]

Next, let's denote the roots of the quadratic equation as \( \alpha \) and \( \beta \). Since the problem states that one of the roots is an integer, we know that \( \alpha \) or \( \beta \) is an integer.

By Vieta's formulas, the sum and product of the roots are given by:

\[
\alpha + \beta = \sqrt{3} + 1 \quad \text{(sum of the roots)}
\]
\[
\alpha \beta = \frac{\sqrt{3}k - 24}{4} \quad \text{(product of the roots)}
\]

Since \( \alpha \) or \( \beta \) is an integer, let's assume \( \alpha \) is an integer. Then \( \beta = \sqrt{3} + 1 - \alpha \).

Now let's substitute \( \beta \) into the product formula:

\[
\alpha(\sqrt{3} + 1 - \alpha) = \frac{\sqrt{3}k - 24}{4}
\]

Expanding the left side:

\[
\alpha(\sqrt{3} + 1) - \alpha^2 = \frac{\sqrt{3}k - 24}{4}
\]

Multiply both sides by 4 to clear the fraction:

\[
4\alpha(\sqrt{3} + 1) - 4\alpha^2 = \sqrt{3}k - 24
\]

Next, isolate \( \sqrt{3}k \):

\[
\sqrt{3}k = 4\alpha(\sqrt{3} + 1) - 4\alpha^2 + 24
\]

To isolate \( k \), divide both sides by \( \sqrt{3} \):

\[
k = \frac{4\alpha(\sqrt{3} + 1) - 4\alpha^2 + 24}{\sqrt{3}}
\]

This equation should yield an integer value for \( k \). We now need to check values of \( \alpha \) (the integer root) that satisfy this condition.

1. \( \alpha = 0 \):

\[
k = \frac{24}{\sqrt{3}} = 8\sqrt{3} \quad \text{(Not an integer)}
\]

2. \( \alpha = 1 \):

\[
k = \frac{4(1)(\sqrt{3} + 1) - 4(1)^2 + 24}{\sqrt{3}} = \frac{4\sqrt{3} + 4 - 4 + 24}{\sqrt{3}} = \frac{4\sqrt{3} + 24}{\sqrt{3}} = 4 + 8 = 12 \quad \text{(integer)}
\]

3. \( \alpha = -1 \):

\[
k = \frac{4(-1)(\sqrt{3} + 1) - 4(-1)^2 + 24}{\sqrt{3}} = \frac{-4\sqrt{3} - 4 - 4 + 24}{\sqrt{3}} = \frac{-4\sqrt{3} + 16}{\sqrt{3}} = -4 + \frac{16}{\sqrt{3}} \quad \text{(Not an integer)}
\]

Thus, \( k = 12 \) is the only valid integer solution.

Therefore, the sum of the values of \( k \) is **12**.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. What happens to the equation if \( k \) is not an integer?
2. Can you solve the quadratic equation when \( k = 12 \)?
3. What is the discriminant of the quadratic equation when \( k = 12 \)?
4. What are the roots of the quadratic equation when \( k = 12 \)?
5. How does the sum and product of roots relate to \( k \) for non-integer values of \( \alpha \)?

**Tip:** When dealing with quadratic equations, always consider the discriminant to determine the nature of the roots (real, complex, or equal).

Explore the solution to a quadratic equation problem where one root is an integer. Understand how Vieta's formulas apply and determine the integer value of 'k'. Suitable for advanced high school students.

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