Let a, b, c be the sides of a triangle where a ≠ b ≠ c and λ ∈ R. If the roots of the equation x² +

Q: Wait, why can we just divide by $(ab+bc+ca)$? What if it's zero?

Since $a, b, c$ are sides of a triangle, they are strictly positive. Thus, their pairwise products are also positive, ensuring we can divide without flipping the inequality. Always check the sign of your divisor!

Q: Why is the inequality strict ($<$) in the geometric part?

Because the problem states $a \neq b \neq c$, which implies a non-degenerate triangle. In a degenerate triangle, the sum of two sides equals the third, but here we have strict inequality, which is why we use the strict less-than sign.

Q: Can I use the AM-GM inequality instead of the triangle inequality?

While AM-GM is powerful, it provides a lower bound ($a^2+b^2+c^2 \ge ab+bc+ca$). Here, we specifically need an upper bound to constrain $\lambda$, which only the triangle inequality provides. Choosing the right tool is half the battle!

Question

Let $a, b, c$ be the sides of a triangle where $a 
eq b 
eq c$ and $\lambda \in R$. If the roots of the equation $x^2 + 2(a+b+c)x + 3\lambda(ab+bc+ca) = 0$ are real, then

Options:

(A) $\lambda < 4/3$

(B) $\lambda > 5/3$

(C) $\lambda \in (1/3, 5/3)$

(D) $\lambda \in (4/3, 5/3)$

Sigma Percentile · Accepted Answer

### Analyzing the Algebraic Gatekeeper

We begin with the quadratic equation:
$$x^2 + 2(a+b+c)x + 3\lambda(ab+bc+ca) = 0$$

For the roots of this equation to be real, the discriminant $D$ must satisfy the condition $D \ge 0$. Identifying the coefficients as $A = 1$, $B = 2(a+b+c)$, and $C = 3\lambda(ab+bc+ca)$, we apply the formula $D = B^2 - 4AC$:

$$[2(a+b+c)]^2 - 4(1)[3\lambda(ab+bc+ca)] \ge 0$$

Dividing by 4 and expanding the square, we obtain:
$$(a+b+c)^2 - 3\lambda(ab+bc+ca) \ge 0$$
$$a^2+b^2+c^2 + 2(ab+bc+ca) - 3\lambda(ab+bc+ca) \ge 0$$

Grouping the terms leads us to our primary algebraic constraint:
$$a^2+b^2+c^2 \ge (3\lambda - 2)(ab+bc+ca)$$

### The Geometric Insight

The variables $a, b,$ and $c$ represent the sides of a triangle. This invokes the Triangle Inequality, which states that the sum of any two sides must be strictly greater than the third:
$a < b+c$, $b < a+c$, and $c < a+b$.

To relate this to our algebraic expression, we multiply these inequalities by $a, b,$ and $c$ respectively:
$a^2 < ab+ac$
$b^2 < ab+bc$
$c^2 < ac+bc$

Summing these three inequalities yields a powerful geometric bound:
$$a^2+b^2+c^2 < 2(ab+bc+ca)$$

### The Final Synthesis

We now combine our algebraic lower bound and our geometric upper bound into a single chain of inequalities:
$$(3\lambda - 2)(ab+bc+ca) \le a^2+b^2+c^2 < 2(ab+bc+ca)$$

Since $a, b,$ and $c$ are side lengths of a triangle, they are strictly positive. Therefore, the term $(ab+bc+ca)$ is positive, allowing us to divide the inequality by this term without reversing the direction:

$$3\lambda - 2 < 2$$

Solving for $\lambda$, we add 2 to both sides to get $3\lambda < 4$. The final range for the parameter is:

$$\lambda < \frac{4}{3}$$

Animated Solution for Mathematics - Quadratic Equations: Let $a, b, c$ be the sides of a triangle where $a \neq = b \neq = c$ and $λ \in R$ . If the roots of the equation $x^{2} + 2 (a + b + c) x + 3 λ (ab + b c + c a) = 0$ are real, then

Select Answer:

Visualized Solution

The Given Setup

Condition for Real Roots

Setting up the Discriminant

Simplifying the Inequality

Expanding the Square

Grouping Terms

The Triangle Inequality

Generating Squared Terms

Summing the Inequalities

Combining the Conditions

Solving for $λ$

The Sigma Insight: Nature of Roots

Analyzing the Algebraic Gatekeeper

The Geometric Insight

The Final Synthesis

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