Solving $x^4 - 21x^2 + 68 = 0$ A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of biquadratic equations. These equations might seem intimidating at first, but with the right approach, they're totally solvable. We'll take a step-by-step look at how to tackle the equation $x^4 - 21x^2 + 68 = 0$. We'll break it down, make it less scary, and find those solutions together. So, let's jump right in and make some math magic happen!

Understanding Biquadratic Equations

Before we dive into the specifics of our equation, let's get a handle on what biquadratic equations actually are. In essence, a biquadratic equation is a quartic equation (an equation with the highest power of the variable being 4) that can be expressed in a specific form. The general form looks like this: $ax^4 + bx^2 + c = 0$, where 'a', 'b', and 'c' are constants, and 'x' is our variable. Notice the key characteristic here: we only have even powers of 'x'. This is what makes it biquadratic, or "quadratic in x-squared", allowing us to use a clever substitution method to simplify the problem. Understanding this fundamental structure is the first key step in solving these equations. Spotting that even powers pattern is like unlocking a secret door to an easier solution process. Once you recognize the form, you're already halfway there!

In our particular case, the equation $x^4 - 21x^2 + 68 = 0$ perfectly fits this mold. We've got $x^4$, an $x^2$ term, and a constant term, which means we can apply the techniques for solving biquadratic equations. This is great news because it transforms a seemingly complex fourth-degree equation into a more manageable quadratic one. This method showcases the power of pattern recognition in mathematics. By seeing the underlying structure, we can choose the most effective tool for the job. So, keep an eye out for this form – it's your signal to use the substitution trick we're about to explore. With this foundation, we're well-prepared to convert our biquadratic equation into a solvable quadratic equation.

The Substitution Trick: Transforming the Equation

The real magic in solving biquadratic equations lies in a simple yet powerful technique: substitution. This trick allows us to transform our fourth-degree equation into a quadratic equation, which we already know how to solve. The idea is to introduce a new variable, let's call it 'u', and make the substitution $u = x^2$. This single step is the key to simplifying the entire problem. By replacing every instance of $x^2$ with 'u', we effectively reduce the complexity of the equation.

So, let's apply this to our equation $x^4 - 21x^2 + 68 = 0$. Notice that $x^4$ can be rewritten as $(x^2)^2$. Now, if we substitute $u = x^2$, then $(x^2)^2$ becomes $u^2$. Similarly, the term $-21x^2$ becomes $-21u$. And the constant term, +68, remains unchanged. Putting it all together, our original equation transforms into $u^2 - 21u + 68 = 0$. Ta-da! We've successfully converted a biquadratic equation into a quadratic equation in terms of 'u'. This is a significant step forward because we now have a familiar form that we can easily solve using standard methods, such as factoring, completing the square, or using the quadratic formula. The substitution technique highlights the beauty of mathematical manipulation, where a clever change of variables can dramatically simplify a problem. By recognizing the structure of the biquadratic equation and applying this substitution, we've paved the way for finding the solutions. Now, let's move on to solving this new quadratic equation and then backtrack to find the values of 'x'.

Solving the Quadratic Equation

Now that we've transformed our biquadratic equation into the quadratic equation $u^2 - 21u + 68 = 0$, it's time to roll up our sleeves and solve for 'u'. There are several methods we can use, but factoring is often the quickest and most straightforward approach if it's possible. So, let's see if we can factor this quadratic equation. Factoring involves finding two numbers that multiply to give the constant term (68 in this case) and add up to give the coefficient of the 'u' term (-21). This may sound tricky, but with a little thought, we can find the right pair.

The factors of 68 are: 1 and 68, 2 and 34, 4 and 17. We need a pair that adds up to -21. Since the constant term is positive and the middle term is negative, we know that both factors must be negative. Looking at our factor pairs, we can see that -4 and -17 fit the bill perfectly: (-4) * (-17) = 68 and (-4) + (-17) = -21. Therefore, we can factor the quadratic equation as $(u - 4)(u - 17) = 0$. Now, to find the values of 'u', we simply set each factor equal to zero and solve. This gives us $u - 4 = 0$ and $u - 17 = 0$, which means $u = 4$ and $u = 17$. Great! We've found the solutions for 'u'. But remember, our original goal was to solve for 'x', so we're not quite done yet. We need to reverse the substitution we made earlier to find the values of 'x'. These values of 'u' are intermediate steps towards our ultimate solution. Once you get comfortable with factoring, you will start to recognize patterns quickly and factor quadratics with confidence. This will speed up your equation-solving abilities significantly. With these 'u' values in hand, we are ready to go back to our original variable 'x' and find the solutions to our biquadratic equation.

Back to $x$: Finding the Solutions

We've successfully solved for 'u' in our transformed quadratic equation, finding that $u = 4$ and $u = 17$. But remember, we're ultimately interested in the values of 'x' that satisfy the original biquadratic equation. To get back to 'x', we need to reverse the substitution we made earlier, which was $u = x^2$. This means we'll substitute our 'u' values back into this equation and solve for 'x'. This step is crucial because it connects our intermediate solutions in terms of 'u' to the actual solutions of the original equation in terms of 'x'. It's like retracing our steps to get back to where we started, but with the added knowledge of the 'u' values we've discovered.

Let's start with $u = 4$. Substituting this into $u = x^2$, we get $x^2 = 4$. To solve for 'x', we take the square root of both sides. Remember that when taking the square root, we need to consider both the positive and negative solutions, because both $2^2$ and $(-2)^2$ equal 4. So, from $x^2 = 4$, we get $x = extit{±}2$. This gives us two solutions: $x = 2$ and $x = -2$. Now, let's do the same for $u = 17$. Substituting this into $u = x^2$, we get $x^2 = 17$. Again, we take the square root of both sides, remembering to consider both positive and negative solutions. This gives us $x = extit{±}\sqrt{17}$. So, we have two more solutions: $x = \sqrt{17}$ and $x = -\sqrt{17}$. Putting it all together, we've found four solutions for our biquadratic equation: $x = 2$, $x = -2$, $x = \sqrt{17}$, and $x = -\sqrt{17}$. These are the values of 'x' that make the original equation $x^4 - 21x^2 + 68 = 0$ true. This process of backtracking and solving for the original variable is a common technique in many mathematical problems. It's a testament to the power of substitution and the importance of carefully retracing your steps to arrive at the final answer. We started with a seemingly complex fourth-degree equation, and by using substitution and our knowledge of quadratic equations, we've successfully found all its solutions.

Summarizing the Solutions

Alright, we've done the hard work and found all the solutions to our biquadratic equation $x^4 - 21x^2 + 68 = 0$. Let's take a moment to summarize our findings and make sure everything is crystal clear. Remember, we started by recognizing the biquadratic form of the equation, which allowed us to use the substitution $u = x^2$. This transformed our equation into a manageable quadratic equation: $u^2 - 21u + 68 = 0$.

We then solved this quadratic equation for 'u', finding two solutions: $u = 4$ and $u = 17$. But we weren't done yet! We needed to go back to our original variable, 'x', by reversing the substitution. This meant substituting our 'u' values back into the equation $u = x^2$ and solving for 'x'. For $u = 4$, we found $x = extit{±}2$, giving us the solutions $x = 2$ and $x = -2$. And for $u = 17$, we found $x = extit{±}\sqrt{17}$, giving us the solutions $x = \sqrt{17}$ and $x = -\sqrt{17}$. So, in total, we have four solutions: $x = 2$, $x = -2$, $x = \sqrt{17}$, and $x = -\sqrt{17}$. These are the values of 'x' that satisfy our original biquadratic equation. It's always a good idea to summarize your solutions like this, especially in more complex problems. It helps you keep track of what you've found and ensures that you haven't missed anything. Plus, it provides a clear and concise answer to the problem. By clearly stating the solutions, we wrap up the problem neatly and demonstrate a complete understanding of the process. With practice, you'll become a pro at solving biquadratic equations and confidently finding all the solutions.

Key Takeaways and Further Practice

So, what have we learned today, guys? We've successfully navigated the world of biquadratic equations and discovered a powerful technique for solving them. The key takeaway here is the substitution method. By recognizing the biquadratic form and using the substitution $u = x^2$, we can transform a seemingly complex fourth-degree equation into a much simpler quadratic equation. This is a testament to the power of algebraic manipulation and the importance of recognizing patterns in mathematics. Remember, a biquadratic equation has the form $ax^4 + bx^2 + c = 0$. The substitution $u = x^2$ turns this into $au^2 + bu + c = 0$, which you can then solve using factoring, the quadratic formula, or completing the square.

Once you've found the solutions for 'u', don't forget to substitute back to find the solutions for 'x'. This involves solving equations of the form $x^2 = u$, which means taking the square root of both sides and remembering to consider both positive and negative solutions. To really master this technique, it's essential to practice. Try solving other biquadratic equations on your own. You can find examples in textbooks, online resources, or even create your own. The more you practice, the more comfortable you'll become with the process, and the faster you'll be able to solve these types of equations. Look for equations with different coefficients and constant terms, and try factoring them, using the quadratic formula, and completing the square. This will help you develop a well-rounded skillset for solving quadratic equations, which is a valuable tool in many areas of mathematics and science. Remember, mathematics is like learning a new language. The more you use it, the more fluent you'll become. So, keep practicing, keep exploring, and keep solving! And who knows, maybe you'll even start seeing biquadratic equations in your dreams!

Correct Substitution:

The correct substitution to write the given equation $x^4 - 21x^2 + 68 = 0$ in quadratic form is:

$u^2 - 21u + 68 = 0$