To find the maximum value of $k$ such that the quadratic equation

$(k+8)x^2 + (k-2)x - 3 = 0$

has real roots, we need to ensure that the discriminant of the equation is non-negative.

Step 1: Identify the Discriminant Condition for Real Roots

For a quadratic equation of the form $ax^2 + bx + c = 0$, the discriminant is given by:

$\Delta = b^2 - 4ac$

The roots are real if and only if $\Delta \geq 0$.

Step 2: Define $a$, $b$, and $c$

In our equation:

• $a = k + 8$
• $b = k - 2$
• $c = -3$

Step 3: Compute the Discriminant

Substitute these values into the discriminant formula:

$\Delta = (k - 2)^2 - 4(k + 8)(-3)$

Expanding this:

$\Delta = (k - 2)^2 + 12(k + 8)$

Calculate each term:

1. $(k - 2)^2 = k^2 - 4k + 4$
2. $12(k + 8) = 12k + 96$

So,

$\Delta = k^2 - 4k + 4 + 12k + 96$

Combine like terms:

$\Delta = k^2 + 8k + 100$

Step 4: Ensure the Discriminant is Non-negative

For the quadratic equation to have real roots, we need:

$k^2 + 8k + 100 \geq 0$

Step 5: Analyze the Quadratic Expression

The expression $k^2 + 8k + 100$ is always positive for all values of $k$ since its discriminant ($8^2 - 4 \cdot 1 \cdot 100$) is negative, indicating it has no real roots and opens upwards.

Therefore, no upper limit on $k$ is needed for real roots to exist. The value of $k$ can be any positive integer.

Conclusion

The maximum value of $k$ such that $(k+8)x^2 + (k-2)x - 3 = 0$ has real roots is unbounded, meaning $k$ can be any positive integer.

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Would you like further clarification on any part of the solution?

Related Questions:

1. What if $k$ were negative? How would the result change?
2. How would we find the roots if $k$ were a specific value, like 1 or 2?
3. Could there be any conditions on $k$ for complex roots to occur?
4. What if we were asked for integer solutions for $x$ given this quadratic?
5. How would this change if we maximized $k$ for complex roots instead?

Tip: For any quadratic equation, always check the discriminant to determine the nature of its roots.