Znalezienie Parametru 'm' W Równaniu Kwadratowym: Warunek X₁ < 1 < X₂

Hey guys! Today, we're diving into a math problem that might seem a bit tricky at first, but we'll break it down step by step to make it super clear. We're dealing with a quadratic equation, and our mission is to find the values of the parameter 'm' that make this equation tick under specific conditions. Specifically, we want the equation to have two roots (solutions), let's call them x₁ and x₂, and these roots need to fall on opposite sides of the number 1 – meaning x₁ is less than 1, and x₂ is greater than 1. Sounds like fun, right? Let's get started!

Rozważania Wstępne i Warunki Brzegowe

Alright, before we jump into the deep end, let's lay down some groundwork. Our equation is (m² + m)x² + 2(m² + m)x + m² - 1 = 0. The first thing we need to consider is that this is a quadratic equation, but not always! When m² + m = 0, the x² term vanishes, and we no longer have a quadratic equation. So, we need to handle this case separately. The equation becomes linear, and we have to solve it differently.

Let's first find the values of m that make the coefficient of x² equal to zero.

m² + m = 0 m(m + 1) = 0

This gives us two possibilities: m = 0 and m = -1. For these values, we will address our given quadratic equation in detail to assess its compliance with the specified criterion of having roots x₁ and x₂ such that x₁ < 1 < x₂. In our discussion, when m = 0, we're left with the equation 0x² + 0x - 1 = 0, which simplifies to -1 = 0. This is clearly not true, meaning there are no solutions when m = 0, so the condition isn't met.

Now, let's explore m = -1. Substituting m into the original equation, we get:

0x² + 0x + 0 = 0.

This equation is true for all values of x. The problem statement, however, requires two distinct roots, which is not the case here. Therefore, the constraint x₁ < 1 < x₂ is not achieved for m = -1 either.

Now that we've considered the cases where the quadratic term disappears, let's move on to the scenarios where the equation is, in fact, quadratic. We must find the m values that not only provide us with two distinct roots but also fulfill our main condition: that one root is less than 1 and the other is greater than 1. This requires us to use the discriminant and the properties of quadratic functions to fulfill the specific conditions of this exercise.

Analiza Discriminantu i Warunek Istnienia Dwóch Pierwiastków

Okay, now let's talk about the discriminant, denoted by Δ (delta). In a quadratic equation of the form ax² + bx + c = 0, the discriminant is calculated as Δ = b² - 4ac. The discriminant tells us about the nature of the roots. If Δ > 0, we have two distinct real roots; if Δ = 0, we have one real root (a repeated root); and if Δ < 0, we have no real roots (the roots are complex).

Since we want two distinct roots (x₁ and x₂), we need Δ > 0. Let's calculate the discriminant for our equation: (m² + m)x² + 2(m² + m)x + m² - 1 = 0. In this case, a = m² + m, b = 2(m² + m), and c = m² - 1. Thus, the discriminant is:

Δ = [2(m² + m)]² - 4(m² + m)(m² - 1) Δ = 4(m² + m)² - 4(m² + m)(m² - 1)

To make our lives easier, we can factor out a 4(m² + m):

Δ = 4(m² + m)[(m² + m) - (m² - 1)] Δ = 4(m² + m)(m² + m - m² + 1) Δ = 4(m² + m)(m + 1)

Now, for Δ > 0, we need 4(m² + m)(m + 1) > 0. Let's further simplify and find the critical points:

4m(m + 1)(m + 1) > 0 4m(m + 1)² > 0

The factor (m + 1)² is always non-negative. This expression is greater than zero when m ≠ -1. So, we need m > 0, because only then is m(m+1)² > 0. Therefore, our condition for two distinct roots is m > 0. We've established an initial boundary, but the main criteria, x₁ < 1 < x₂, remains to be addressed.

Warunek x₁ < 1 < x₂ – Kluczowy Moment

Alright, here comes the heart of the matter. We want the roots to sandwich the number 1. This means one root is less than 1, and the other is greater than 1. We can use a neat trick to achieve this without directly calculating the roots. We will use a property of quadratic functions:

If a quadratic function f(x) = ax² + bx + c has roots x₁ and x₂, then f(x) changes its sign between x₁ and x₂. If f(1) has the opposite sign to 'a', then this guarantees that one root is less than 1, and the other is greater than 1 (or vice versa).

In our case, a = m² + m and f(x) = (m² + m)x² + 2(m² + m)x + m² - 1. So, we'll calculate f(1) and ensure it has the opposite sign to a.

f(1) = (m² + m)(1)² + 2(m² + m)(1) + m² - 1 f(1) = m² + m + 2m² + 2m + m² - 1 f(1) = 4m² + 3m - 1

Now, we need to make sure that (m² + m) and (4m² + 3m - 1) have opposite signs. That is:

(m² + m)(4m² + 3m - 1) < 0

Let's factorize 4m² + 3m - 1, we get (m + 1)(4m - 1). This yields:

m(m + 1)(m + 1)(4m - 1) < 0

To analyze this, we need to find the zeros of the expression. They are m = 0, m = -1, and m = 1/4. Now we build the sign chart.

So, the inequality is satisfied for 0 < m < 1/4. We already established that m > 0 for two distinct roots to exist. Therefore, combining this with the previous result, we see that the solution to our problem is 0 < m < 1/4. We've cracked it!

Podsumowanie i Ostateczna Odpowiedź

To sum up, we started with a quadratic equation, looked at the conditions for the existence of two roots, used the discriminant to ensure they were distinct, and then applied a clever trick using f(1) and the sign of 'a' to ensure that the roots were on either side of 1. By doing all that, we managed to pin down the correct values of m.

Therefore, the values of the parameter m that satisfy the given condition x₁ < 1 < x₂ are those that fall within the range 0 < m < 1/4.

This means that the values of m must be greater than zero but less than one-fourth for the equation to have two roots that meet our requirements. Congratulations, guys, we've solved it!

In short, we looked at how to solve quadratic equations, discriminant, and analysis of roots within a specific range. This is a fun example of how different tools in math come together to solve a specific problem. I hope it helps and happy studying!

Key Takeaways:

• Understanding the Discriminant: Knowing how the discriminant helps determine the nature of the roots (real, distinct, etc.) is crucial.
• The Sign of f(1): Using f(1) to determine the position of the roots relative to 1 is a powerful trick.
• Combining Conditions: Combining multiple conditions (Δ > 0 and the sign of f(1)) is a common strategy in these types of problems.

This exploration highlights the importance of combining several math concepts to find solutions. Remember to keep practicing and try more problems to master these skills. Cheers! And see you in the next lesson!Remember that understanding the core concepts and practicing are the keys to acing these types of problems. Good luck and have fun!