Quadratic Equation Roots: Finding M Values

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Hey guys! Let's dive into a fun math problem today that involves quadratic equations and figuring out when their roots have a special relationship. We're going to explore the equation x² + (3m - 2)x + m + 2 = 0 and determine the values of the parameter 'm' that make twice the product of the roots equal to their sum. Sounds like a puzzle, right? Let's break it down step by step!

Understanding the Problem: Roots and Relationships

So, what exactly are we trying to solve? The heart of this problem lies in understanding the connection between the coefficients of a quadratic equation and the properties of its roots. Remember, the roots of a quadratic equation are the values of 'x' that make the equation true (i.e., where the parabola crosses the x-axis). When we say we have two roots, it means there are two such 'x' values. We're looking for situations where a specific condition is met: twice the product of these roots is equal to their sum. To tackle this, we'll use some handy formulas derived from Vieta's formulas.

Vieta's formulas are our secret weapon here. They provide a direct link between the coefficients of a polynomial (in our case, a quadratic) and the sums and products of its roots. For a general quadratic equation ax² + bx + c = 0, Vieta's formulas tell us:

  • The sum of the roots (x₁ + x₂) = -b/a
  • The product of the roots (x₁ * x₂) = c/a

In our specific equation, x² + (3m - 2)x + m + 2 = 0, we can identify:

  • a = 1
  • b = (3m - 2)
  • c = (m + 2)

Therefore, using Vieta's formulas, we have:

  • x₁ + x₂ = -(3m - 2) / 1 = -3m + 2
  • x₁ * x₂ = (m + 2) / 1 = m + 2

Now, let's get back to the condition we need to satisfy: 2(x₁ * x₂) = x₁ + x₂. This equation is the key to finding the values of 'm'.

Setting Up the Equation and Solving for 'm'

Alright, we've got all the pieces of the puzzle. We know the relationship between the roots (twice the product equals the sum), and we have expressions for the sum and product of the roots in terms of 'm'. Let's put them together! The key here is to carefully substitute our expressions from Vieta's formulas into the given condition. This will create an equation that we can solve for 'm'.

Our condition is: 2(x₁ * x₂) = x₁ + x₂

Substituting the expressions we derived from Vieta's formulas, we get:

  • 2(m + 2) = -3m + 2

Now we have a simple algebraic equation in terms of 'm'. Let's solve it:

  1. Expand the left side: 2m + 4 = -3m + 2
  2. Add 3m to both sides: 5m + 4 = 2
  3. Subtract 4 from both sides: 5m = -2
  4. Divide both sides by 5: m = -2/5

So, it looks like we have a potential solution for 'm'. But hold on, we're not quite done yet! Remember, for a quadratic equation to have two roots (which is a requirement of the problem), its discriminant must be greater than zero. We need to check this condition.

Checking the Discriminant: Ensuring Two Roots

Before we can confidently say that m = -2/5 is the correct answer, we need to make sure that this value of 'm' actually results in the quadratic equation having two distinct real roots. This is where the discriminant comes in. The discriminant, often denoted by Δ (Delta), is a part of the quadratic formula that tells us about the nature of the roots. For a quadratic equation ax² + bx + c = 0, the discriminant is given by:

  • Δ = b² - 4ac

Here's how the discriminant helps us:

  • If Δ > 0, the equation has two distinct real roots.
  • If Δ = 0, the equation has one real root (a repeated root).
  • If Δ < 0, the equation has no real roots (two complex roots).

Since our problem specifies that we need two roots, we need to ensure that Δ > 0. Let's calculate the discriminant for our equation, x² + (3m - 2)x + m + 2 = 0:

  • Δ = (3m - 2)² - 4 * 1 * (m + 2)

Now, substitute m = -2/5 into this expression and see if the discriminant is positive:

  • Δ = (3(-2/5) - 2)² - 4(-2/5 + 2)
  • Δ = (-6/5 - 10/5)² - 4(-2/5 + 10/5)
  • Δ = (-16/5)² - 4(8/5)
  • Δ = 256/25 - 32/5
  • Δ = 256/25 - 160/25
  • Δ = 96/25

Great! Since 96/25 is a positive number, the discriminant is greater than zero when m = -2/5. This confirms that our quadratic equation has two distinct real roots for this value of 'm'.

The Final Answer: m = -2/5

Phew! We made it through all the steps. We started by understanding the problem and Vieta's formulas, then we set up an equation based on the given condition and solved for 'm'. Finally, and very importantly, we checked the discriminant to make sure our solution actually resulted in two distinct real roots.

Therefore, the value of the parameter 'm' for which the equation x² + (3m - 2)x + m + 2 = 0 has two roots such that twice their product equals their sum is m = -2/5.

This type of problem highlights the beautiful connections between different parts of mathematics. By understanding the relationship between the coefficients and roots of a polynomial, we can solve some pretty cool problems! Keep practicing, guys, and you'll become quadratic equation masters in no time!