Solving For K: Root Difference In Quadratic Equations

by Dimemap Team 54 views

Hey guys! Let's dive into a fun algebra problem where we need to figure out the value of 'k' in different quadratic equations. The twist? We know that the difference between the roots (x1 and x2) of these equations is 1. Sounds interesting, right? We'll break down each equation step by step, so you can follow along easily. So, grab your thinking caps, and let's get started!

Understanding the Basics: Roots and Quadratic Equations

Before we jump into solving, let's quickly recap some key concepts. A quadratic equation is generally in the form ax² + bx + c = 0. The roots (x1 and x2) are the values of x that satisfy this equation. There are a couple of super handy formulas that relate the roots to the coefficients of the equation:

  • Sum of roots (x1 + x2) = -b/a
  • Product of roots (x1 * x2) = c/a

These formulas will be our best friends in solving for 'k'. Also, we have the condition given: |x1 - x2| = 1. This means the absolute difference between the roots is 1. We're going to use this, along with the sum and product formulas, to crack each equation.

Equation 1: 5x² + kx + 1 = 0

Okay, let's tackle the first equation: 5x² + kx + 1 = 0. Here, a = 5, b = k, and c = 1. Using our root formulas, we get:

  • x1 + x2 = -k/5
  • x1 * x2 = 1/5

Now, we know that |x1 - x2| = 1. To make things easier, let's square both sides: (x1 - x2)² = 1. Expanding this, we get x1² - 2x1x2 + x2² = 1. We can rewrite this as (x1 + x2)² - 4x1x2 = 1. Why did we do this? Because we know the values of (x1 + x2) and (x1 * x2) in terms of 'k'!

Substituting the values, we have (-k/5)² - 4(1/5) = 1. Simplifying, we get k²/25 - 4/5 = 1. Let's solve for k²: k²/25 = 1 + 4/5 = 9/5. Therefore, k² = (9/5) * 25 = 45. Taking the square root, we find k = ±√45 = ±3√5. So, for the first equation, k can be either 3√5 or -3√5. Great job so far, guys!

Equation 2: 5x² - kx - 1 = 0

Next up, we have 5x² - kx - 1 = 0. This time, a = 5, b = -k, and c = -1. Applying the root formulas:

  • x1 + x2 = -(-k)/5 = k/5
  • x1 * x2 = -1/5

We still have the condition |x1 - x2| = 1, so (x1 - x2)² = 1. Again, we rewrite it as (x1 + x2)² - 4x1x2 = 1. Substituting the values in terms of 'k', we get (k/5)² - 4(-1/5) = 1. This simplifies to k²/25 + 4/5 = 1. Let's isolate k²: k²/25 = 1 - 4/5 = 1/5. Thus, k² = (1/5) * 25 = 5. Taking the square root gives us k = ±√5. So, for the second equation, k can be either √5 or -√5. You're doing awesome, keep going!

Equation 3: kx² - 6x + 4 = 0

Now, let's look at kx² - 6x + 4 = 0. This one is a bit trickier because 'k' is also the coefficient of x². Here, a = k, b = -6, and c = 4. The root formulas give us:

  • x1 + x2 = -(-6)/k = 6/k
  • x1 * x2 = 4/k

We still have (x1 - x2)² = 1, which we rewrite as (x1 + x2)² - 4x1x2 = 1. Substituting, we get (6/k)² - 4(4/k) = 1. This simplifies to 36/k² - 16/k = 1. To get rid of the fractions, multiply everything by k²: 36 - 16k = k². Rearranging, we have a quadratic equation in terms of k: k² + 16k - 36 = 0.

We can solve this quadratic equation for k using the quadratic formula: k = [-b ± √(b² - 4ac)] / (2a). Here, a = 1, b = 16, and c = -36. Plugging in the values, we get:

k = [-16 ± √(16² - 4 * 1 * -36)] / (2 * 1) k = [-16 ± √(256 + 144)] / 2 k = [-16 ± √400] / 2 k = [-16 ± 20] / 2

So, k = (-16 + 20) / 2 = 2 or k = (-16 - 20) / 2 = -18. Thus, for the third equation, k can be 2 or -18. Fantastic work!

Equation 4: 4x² - kx + 3 = 0

Last but not least, let's solve 4x² - kx + 3 = 0. In this equation, a = 4, b = -k, and c = 3. Using the root formulas:

  • x1 + x2 = -(-k)/4 = k/4
  • x1 * x2 = 3/4

Again, we use (x1 - x2)² = 1, rewritten as (x1 + x2)² - 4x1x2 = 1. Substituting the values, we get (k/4)² - 4(3/4) = 1. This simplifies to k²/16 - 3 = 1. Let's solve for k²: k²/16 = 1 + 3 = 4. Therefore, k² = 4 * 16 = 64. Taking the square root, we find k = ±√64 = ±8. So, for the fourth equation, k can be either 8 or -8. You've nailed it!

Final Thoughts and Key Takeaways

Wow, guys, we've solved for 'k' in four different quadratic equations! That's quite an achievement. Let's quickly summarize what we've found:

  • For 5x² + kx + 1 = 0, k = ±3√5
  • For 5x² - kx - 1 = 0, k = ±√5
  • For kx² - 6x + 4 = 0, k = 2 or -18
  • For 4x² - kx + 3 = 0, k = ±8

The key to solving these problems was using the relationships between the roots and the coefficients of the quadratic equations, along with the given condition |x1 - x2| = 1. We transformed the difference of roots into a more usable form by squaring it and relating it to the sum and product of roots.

Remember, algebra might seem daunting at first, but with practice and a good understanding of the fundamentals, you can conquer any problem. Keep practicing, keep exploring, and most importantly, have fun with math! You've got this! If you found this breakdown helpful, give it a thumbs up and share it with your friends. Let's spread the algebra love! Until next time, keep those brains buzzing!