Key Concepts and Formulas

• Substitution: Replacing a complex expression with a single variable to simplify equations.
• Factoring Quadratic Expressions: Expressing a quadratic expression as a product of two linear expressions.
• Discriminant of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. The equation has real roots if and only if $D \ge 0$.
• Solving Quadratic Inequalities: Finding the intervals where a quadratic expression is positive, negative, or zero.

Step-by-Step Solution

Step 1: Simplify the Equation using Substitution

We are given the equation: $(3x^2 + 4x + 3)^2 - (k + 1)(3x^2 + 4x + 3)(3x^2 + 4x + 2) + k(3x^2 + 4x + 2)^2 = 0$ To simplify this equation, we introduce new variables: Let $P = 3x^2 + 4x + 3$ and $Q = 3x^2 + 4x + 2$.

Why this step? This substitution makes the equation easier to manipulate algebraically by reducing its visual complexity. We are essentially grouping repeating terms.

Substituting $P$ and $Q$ into the original equation, we get: $P^2 - (k+1)PQ + kQ^2 = 0$

Step 2: Factor the Transformed Equation

The equation $P^2 - (k+1)PQ + kQ^2 = 0$ is a homogeneous quadratic equation in terms of $P$ and $Q$. We can factor this equation.

Splitting the middle term $-(k+1)PQ$ into $-kPQ - PQ$: $P^2 - kPQ - PQ + kQ^2 = 0$ Now, factor by grouping: $P(P - kQ) - Q(P - kQ) = 0$ $(P - kQ)(P - Q) = 0$

Why this step? Factoring the equation allows us to break the problem down into simpler cases. For the product of two terms to be zero, at least one of the terms must be zero.

Step 3: Analyze the Possible Cases

From the factored form $(P - kQ)(P - Q) = 0$, we have two possibilities:

Case A: $P - Q = 0 \implies P = Q$ Substitute back the expressions for $P$ and $Q$: $3x^2 + 4x + 3 = 3x^2 + 4x + 2$ Simplifying this equation, we get: $3 = 2$

Why this step? We substitute back the original expressions to see what this case implies for $x$. The result $3=2$ is a false statement. Therefore, Case A does not lead to any real roots for the original equation, and we discard it.

Case B: $P - kQ = 0 \implies P = kQ$ Substitute back the expressions for $P$ and $Q$: $3x^2 + 4x + 3 = k(3x^2 + 4x + 2)$

Why this step? This is the only remaining valid relationship derived from the factored equation. We must now determine the values of $k$ for which this equation (which is now in terms of $x$) has real roots.

To apply the discriminant condition, we need to rearrange this equation into the standard quadratic form $Ax^2 + Bx + C = 0$: $3x^2 + 4x + 3 = 3kx^2 + 4kx + 2k$ Move all terms to one side: $(3 - 3k)x^2 + (4 - 4k)x + (3 - 2k) = 0$ Factor out common terms: $3(1 - k)x^2 + 4(1 - k)x + (3 - 2k) = 0$ For convenience, we can multiply the entire equation by $-1$: $3(k - 1)x^2 + 4(k - 1)x + (2k - 3) = 0$

Step 4: Apply the Discriminant Condition for Real Roots

For the quadratic equation $3(k-1)x^2 + 4(k-1)x + (2k-3) = 0$ to have real roots, its discriminant $D$ must be greater than or equal to zero ($D \ge 0$).

Here, the coefficients are: $A = 3(k-1)$ $B = 4(k-1)$ $C = 2k-3$

The discriminant is given by $D = B^2 - 4AC$: $D = (4(k-1))^2 - 4(3(k-1))(2k-3) \ge 0$

Why this step? The discriminant is a direct measure of the nature of the roots of a quadratic equation. $D \ge 0$ ensures that the roots are real.

Now, let's simplify the inequality: $16(k-1)^2 - 12(k-1)(2k-3) \ge 0$ Notice that $4(k-1)$ is a common factor. Factor it out: $4(k-1) [4(k-1) - 3(2k-3)] \ge 0$ Simplify the expression inside the square brackets: $4(k-1) [4k - 4 - 6k + 9] \ge 0$ $4(k-1) [-2k + 5] \ge 0$ To make the leading coefficient of $k$ positive for easier inequality solving, we multiply the term $[-2k+5]$ by $-1$ and compensate by changing the inequality sign. We can also divide by 4: $(k-1)(-(2k - 5)) \ge 0$ $-(k-1)(2k - 5) \ge 0$ Multiply by $-1$ and reverse the inequality sign: $(k-1)(2k - 5) \le 0$

Step 5: Solve the Quadratic Inequality for $k$

We need to solve the inequality $(k-1)(2k-5) \le 0$.

Why this step? Solving this inequality will give us the range of $k$ values for which the discriminant is non-negative, and thus for which the quadratic equation in $x$ has real roots.

To solve this quadratic inequality, we first find the roots of the corresponding quadratic equation $(k-1)(2k-5) = 0$. These roots are $k=1$ and $k=\frac{5}{2}$. These roots divide the number line into three intervals: $(-\infty, 1)$, $[1, 5/2]$, and $(5/2, \infty)$. The expression $f(k) = (k-1)(2k-5)$ represents a parabola that opens upwards. For an upward-opening parabola, $f(k) \le 0$ when $k$ is between its roots.

Therefore, the inequality holds for: $1 \le k \le \frac{5}{2}$ In interval notation, this is $k \in \left[1, \frac{5}{2}\right]$.

Step 6: Address Degenerate Case and Initial Constraint

We have two additional conditions to consider:

1. The problem states that $K > -1$. Our current solution $k \in \left[1, \frac{5}{2}\right]$ fully satisfies this condition, as all values in this interval are greater than $-1$.

2. Degenerate Case: The discriminant condition $D \ge 0$ is valid only for a quadratic equation (i.e., when the coefficient of $x^2$ is non-zero). We need to check what happens if the coefficient $A = 3(k-1)$ becomes zero.

   • If $3(k-1) = 0$, then $k=1$.
   • Let's substitute $k=1$ back into the equation from Case B: $3(1 - 1)x^2 + 4(1 - 1)x + (3 - 2(1)) = 0$ $0x^2 + 0x + (3 - 2) = 0$ $0x + 1 = 0$ $1 = 0$ Why this step? If $A=0$, the equation is no longer quadratic, and the discriminant is irrelevant. In this specific instance, for $k=1$, the equation reduces to $1=0$, which is a contradiction. This implies that for $k=1$, there are no real values of $x$ that can satisfy the equation. Hence, the original equation has no real roots when $k=1$.

   Since $k=1$ does not yield real roots, we must exclude it from our solution set.

Combining the inequality solution $1 \le k \le \frac{5}{2}$ with the exclusion of $k=1$, the final set of values for $k$ is: $1 < k \le \frac{5}{2}$ In interval notation, this is $k \in \left(1, \frac{5}{2}\right]$.

Common Mistakes & Tips

• Forgetting the Degenerate Case: Always check what happens when the coefficient of $x^2$ is zero. The discriminant condition only applies to quadratic equations.
• Incorrectly Solving the Inequality: Be careful when multiplying or dividing inequalities by negative numbers. Remember to flip the inequality sign.
• Not Considering the Initial Constraint: Always check if the solution satisfies the given constraints on the variables.

Summary

We used substitution to simplify the given equation, then factored it into two cases. One case led to a contradiction, and the other led to a quadratic equation in $x$. We then applied the discriminant condition to find the range of $k$ values for which the quadratic equation has real roots. Finally, we checked the degenerate case where the coefficient of $x^2$ is zero and excluded the corresponding value of $k$ from the solution set. The final set of values of $K > -1$ for which the equation has real roots is $\left(1, \frac{5}{2}\right]$.

Final Answer

The final answer is $\boxed{\left(1, \frac{5}{2}\right]}$, which corresponds to option (A).