Solving Biquadratic Equations: Step-by-Step Guide

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Solving Biquadratic Equations: A Step-by-Step Guide

Hey guys! Let's dive into the world of biquadratic equations. If you're scratching your head over equations like 4x⁴ - 13x² + 3 = 0, you're in the right place. These might look intimidating, but don't worry, we'll break them down together. This guide will walk you through solving five different biquadratic equations, making the process super clear and easy to follow. We'll cover everything from the initial setup to finding those final solutions. So, grab your pencils, and let's get started!

What are Biquadratic Equations?

Before we jump into solving, let's quickly understand what biquadratic equations are. Biquadratic equations are essentially quadratic equations in disguise. They have the general form of ax⁴ + bx² + c = 0, where 'a', 'b', and 'c' are constants, and 'x' is our variable. Notice how the exponents are 4 and 2? That’s the key. The trick to solving these is to turn them into simpler quadratic equations using a clever substitution.

The Substitution Method: Our Secret Weapon

The substitution method is our best friend when it comes to biquadratic equations. We introduce a new variable, usually 't', such that t = x². This transforms our biquadratic equation into a standard quadratic equation in terms of 't'. Once we solve for 't', we can easily find the values of 'x' by taking the square root. Sounds simple, right? Let's see it in action!

Solving the Equations: A Detailed Walkthrough

Now, let's tackle those equations one by one. We'll go through each step, so you can see exactly how it's done. Ready? Let's do this!

a) 4x⁴ - 13x² + 3 = 0

First up, we have the equation 4x⁴ - 13x² + 3 = 0. Let’s get this sorted! Here’s how we'll do it:

  1. Substitution: Let t = x². This changes our equation to 4t² - 13t + 3 = 0. See? Much simpler!
  2. Solve the Quadratic Equation: Now we need to solve for 't'. We can use the quadratic formula, factoring, or any method you prefer. For this one, let’s try factoring. We're looking for two numbers that multiply to 4 * 3 = 12 and add up to -13. Those numbers are -12 and -1. So, we rewrite the middle term: 4t² - 12t - t + 3 = 0.
  3. Factor by Grouping: Group the terms and factor: 4t(t - 3) - 1(t - 3) = 0. Now we have (4t - 1)(t - 3) = 0.
  4. Find the Values of t: Set each factor equal to zero: 4t - 1 = 0 gives us t = 1/4, and t - 3 = 0 gives us t = 3.
  5. Back-Substitute: Remember, t = x². So, we need to solve for 'x'. We have two cases: x² = 1/4 and x² = 3.
  6. Solve for x: For x² = 1/4, we get x = ±√(1/4) = ±1/2. For x² = 3, we get x = ±√3.
  7. Final Solutions: So, our solutions for this equation are x = 1/2, x = -1/2, x = √3, and x = -√3. Awesome!

b) 9x⁴ + 17x² - 2 = 0

Next, let’s tackle 9x⁴ + 17x² - 2 = 0. Same strategy applies here:

  1. Substitution: Let t = x². The equation becomes 9t² + 17t - 2 = 0.
  2. Solve the Quadratic Equation: This time, let’s use the quadratic formula. The quadratic formula is t = (-b ± √(b² - 4ac)) / (2a). Here, a = 9, b = 17, and c = -2.
  3. Apply the Formula: Plugging in the values, we get t = (-17 ± √(17² - 4 * 9 * -2)) / (2 * 9) = (-17 ± √(289 + 72)) / 18 = (-17 ± √361) / 18 = (-17 ± 19) / 18.
  4. Find the Values of t: This gives us two values for t: t = (-17 + 19) / 18 = 2 / 18 = 1/9 and t = (-17 - 19) / 18 = -36 / 18 = -2.
  5. Back-Substitute: Again, t = x². So, x² = 1/9 and x² = -2.
  6. Solve for x: For x² = 1/9, we get x = ±√(1/9) = ±1/3. For x² = -2, we get x = ±√(-2) = ±i√2 (where 'i' is the imaginary unit, √-1).
  7. Final Solutions: Our solutions are x = 1/3, x = -1/3, x = i√2, and x = -i√2. Great job!

c) x⁴ + 7x² - 8 = 0

Moving on to x⁴ + 7x² - 8 = 0. You guys are getting the hang of this, right?

  1. Substitution: Let t = x². This gives us t² + 7t - 8 = 0.
  2. Solve the Quadratic Equation: Let's factor this one. We need two numbers that multiply to -8 and add up to 7. Those are 8 and -1. So, we can rewrite the equation as (t + 8)(t - 1) = 0.
  3. Find the Values of t: Setting each factor to zero, we get t + 8 = 0, which means t = -8, and t - 1 = 0, which means t = 1.
  4. Back-Substitute: Remember, t = x². So, x² = -8 and x² = 1.
  5. Solve for x: For x² = -8, we get x = ±√(-8) = ±2i√2. For x² = 1, we get x = ±√1 = ±1.
  6. Final Solutions: Our solutions are x = 2i√2, x = -2i√2, x = 1, and x = -1. Keep it up!

d) 9x⁴ - 19x² + 2 = 0

Now for 9x⁴ - 19x² + 2 = 0. Let's keep the momentum going:

  1. Substitution: Let t = x². Our equation becomes 9t² - 19t + 2 = 0.
  2. Solve the Quadratic Equation: Let's use factoring again. We need two numbers that multiply to 9 * 2 = 18 and add up to -19. Those numbers are -18 and -1. Rewrite the middle term: 9t² - 18t - t + 2 = 0.
  3. Factor by Grouping: Group the terms and factor: 9t(t - 2) - 1(t - 2) = 0. This gives us (9t - 1)(t - 2) = 0.
  4. Find the Values of t: Set each factor equal to zero: 9t - 1 = 0 gives us t = 1/9, and t - 2 = 0 gives us t = 2.
  5. Back-Substitute: Since t = x², we have x² = 1/9 and x² = 2.
  6. Solve for x: For x² = 1/9, we get x = ±√(1/9) = ±1/3. For x² = 2, we get x = ±√2.
  7. Final Solutions: Our solutions are x = 1/3, x = -1/3, x = √2, and x = -√2. Almost there!

f) 9x⁴ + 14x² - 8 = 0

Last but not least, 9x⁴ + 14x² - 8 = 0. Let's finish strong!

  1. Substitution: Let t = x². This gives us 9t² + 14t - 8 = 0.
  2. Solve the Quadratic Equation: Time to factor. We need two numbers that multiply to 9 * -8 = -72 and add up to 14. Those numbers are 18 and -4. Rewrite the middle term: 9t² + 18t - 4t - 8 = 0.
  3. Factor by Grouping: Group the terms and factor: 9t(t + 2) - 4(t + 2) = 0. This gives us (9t - 4)(t + 2) = 0.
  4. Find the Values of t: Set each factor equal to zero: 9t - 4 = 0 gives us t = 4/9, and t + 2 = 0 gives us t = -2.
  5. Back-Substitute: Since t = x², we have x² = 4/9 and x² = -2.
  6. Solve for x: For x² = 4/9, we get x = ±√(4/9) = ±2/3. For x² = -2, we get x = ±√(-2) = ±i√2.
  7. Final Solutions: Our solutions are x = 2/3, x = -2/3, x = i√2, and x = -i√2. You nailed it!

Key Takeaways and Tips for Success

  • Substitution is Key: Always start by substituting t = x² to simplify the equation.
  • Choose Your Method: Whether you prefer factoring, the quadratic formula, or completing the square, pick the method that works best for you.
  • Don't Forget the Plus/Minus: When taking the square root to solve for 'x', remember to consider both the positive and negative roots.
  • Complex Solutions: Be prepared to encounter complex solutions (with 'i') when dealing with negative square roots.

Practice Makes Perfect

Solving biquadratic equations might seem tricky at first, but with practice, you’ll become a pro. Try working through more examples, and don’t hesitate to revisit these steps whenever you need a refresher. Remember, the more you practice, the easier it gets.

Conclusion: You've Got This!

So there you have it! We've walked through solving five different biquadratic equations step-by-step. You've learned the substitution method, tackled factoring and the quadratic formula, and even encountered complex solutions. Keep practicing, and you’ll be solving these equations in your sleep. You've totally got this, guys! If you found this guide helpful, share it with your friends, and keep exploring the amazing world of math!