The set of all real numbers a such that a² + 2a, 2a + 3 and a² + 3a + 8 are the sides of a triangle

Q: Wait, why do we need to check if the sides are positive? Isn't the triangle inequality enough?

That is a fantastic question! While the triangle inequality ensures that three lines can connect to form a triangle, it does not guarantee that the lengths themselves are physically possible. In algebra, an expression like $a^2 + 2a$ could result in a negative number depending on the value of $a$. Since a triangle cannot have a negative side length, we must enforce the positivity constraint ($x > 0, y > 0, z > 0$) to ensure our solutions are physically valid.

Q: What if the discriminant of the quadratic was positive? How would that change the intersection?

If the discriminant $D$ of the quadratic $2a^2 + 3a + 5 > 0$ had been positive, the quadratic would have had real roots. In that case, the inequality $2a^2 + 3a + 5 > 0$ would only be true for specific intervals of $a$ (either between the roots or outside them). We would then have to find the intersection of those intervals with our other conditions ($a > 5$, etc.). Because our discriminant was negative, the quadratic is always positive, which simplifies our life significantly!

Q: Why do we use an open interval $(5, \infty)$ instead of a closed one $[5, \infty)$?

Excellent observation! The triangle inequality theorem states that the sum of two sides must be *strictly* greater than the third side. If the sum were *equal* to the third side, the three points would be collinear, forming a 'degenerate' triangle (essentially a straight line). Since the problem asks for a triangle, we must exclude the case where the sum equals the third side, hence the use of the open interval.

Question

The set of all real numbers $a$ such that $a^2 + 2a$, $2a + 3$ and $a^2 + 3a + 8$ are the sides of a triangle is ..................

Sigma Percentile · Accepted Answer

### The Geometry of Algebra: A Journey into Triangle Existence Welcome, future engineer! Today, we are going to embark on a journey that bridges the gap between pure algebra and the elegant world of geometry. We are given three expressions: $x = a^2 + 2a$, $y = 2a + 3$, and $z = a^2 + 3a + 8$. Our mission is to find the set of all real numbers $a$ such that these expressions can form the sides of a valid triangle. ### The Silent Constraint: The Reality of Length Before we even think about the triangle inequality, we must address the most fundamental reality of geometry: length. In the physical world, a side of a triangle cannot be zero, and it certainly cannot be negative. Therefore, our first step is to ensure that $x$, $y$, and $z$ are all strictly positive. Let us look at $y = 2a + 3 > 0$. This gives us $a > -\frac{3}{2}$. Next, consider $x = a^2 + 2a > 0$. Factoring this, we get $a(a+2) > 0$. This inequality holds when $a > 0$ or $a < -2$. When we intersect these conditions with our requirement for $y$, we find that for all sides to be positive, we must have $a > 0$. This is our first anchor point on the number line. ### The Three Pillars of Triangle Existence Now, we invoke the Triangle Inequality Theorem. For any triangle with sides $x, y, z$, the following three conditions must hold simultaneously: 1. $x + y > z$ 2. $y + z > x$ 3. $z + x > y$ Let us tackle these one by one. First, $x + y > z$ becomes: $$(a^2 + 2a) + (2a + 3) > a^2 + 3a + 8$$ Expanding this, we get $a^2 + 4a + 3 > a^2 + 3a + 8$. Notice the beauty of the algebra here: the $a^2$ terms cancel out perfectly! We are left with $4a + 3 > 3a + 8$, which simplifies to **$a > 5$**. This is a powerful constraint. Next, we examine $y + z > x$. Substituting our expressions, we have: $$(2a + 3) + (a^2 + 3a + 8) > a^2 + 2a$$ Simplifying, we get $a^2 + 5a + 11 > a^2 + 2a$. Again, the $a^2$ terms vanish, leaving $5a + 11 > 2a$, or $3a > -11$, which means $a > -\frac{11}{3}$. ### The Quadratic Mirage Finally, we look at $z + x > y$. This gives us: $$(a^2 + 3a + 8) + (a^2 + 2a) > 2a + 3$$ This simplifies to $2a^2 + 5a + 8 > 2a + 3$, which rearranges to $2a^2 + 3a + 5 > 0$. Let us calculate the discriminant $D = b^2 - 4ac$. Here, $D = 3^2 - 4(2)(5) = 9 - 40 = -31$. Since $D < 0$ and the leading coefficient is positive, this quadratic is always positive for all real $a$. It imposes no further restrictions. ### The Final Intersection We have gathered our conditions: $a > 0$, $a > 5$, and $a > -\frac{11}{3}$. To satisfy all these simultaneously, we must find the intersection of these sets. The most restrictive condition is $a > 5$. Therefore, the set of all real numbers $a$ that allow these expressions to form a triangle is the open interval **$(5, \infty)$**. Take a moment to appreciate this. We started with three abstract algebraic expressions and, through the rigorous application of geometric laws, distilled them into a clear, precise range.

Animated Solution for Mathematics - Trigonometry: The set of all real numbers $a$ such that $a^{2} + 2 a$ , $2 a + 3$ and $a^{2} + 3 a + 8$ are the sides of a triangle is ..................

Visualized Solution

Defining the Triangle Sides

The Triangle Inequality Theorem

Constraint: Positivity of Sides

Solving for Positivity

First Inequality: $x + y > z$

Solving $x + y > z$

Second Inequality: $y + z > x$

Solving $y + z > x$

Third Inequality: $z + x > y$

Solving $z + x > y$

Checking the Quadratic

Finding the Intersection

The Sigma Insight: Properties of Triangles

The Geometry of Algebra

The Silent Constraint

The Three Pillars of Triangle Existence

The Quadratic Mirage

The Final Intersection

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of all real numbers a such that a2+2a, 2a+3 and a2+3a+8 are the sides of a triangle is ..................