Solution Set For The Equation X^4 - 13x^2 + 36 = 0
Hey everyone! Today, we're diving into the fascinating world of algebra to solve a quartic equation. Specifically, we're tackling the equation x^4 - 13x^2 + 36 = 0 and figuring out its solution set within the realm of real numbers. It might look intimidating at first, but trust me, we'll break it down step by step, making it super easy to understand. So, grab your thinking caps, and let's get started!
Understanding the Problem: A Quartic Equation
So, what exactly are we dealing with here? The equation x^4 - 13x^2 + 36 = 0 is a quartic equation, meaning it's a polynomial equation where the highest power of the variable (in this case, 'x') is 4. Quartic equations can have up to four solutions or roots. Our mission is to find those roots, but with a twist: we're only interested in the real number solutions. Real numbers, as a quick reminder, are all the numbers you can find on a number line – including positives, negatives, fractions, decimals, and even the sneaky zero.
Before we jump into solving, it's worth noting that quartic equations can sometimes be tricky beasts. Unlike quadratic equations (where the highest power is 2), there isn't a simple, universally applicable formula to directly find the roots of a quartic equation. However, fear not! This particular quartic equation has a special form that makes it much more manageable. We'll exploit this special form using a clever technique called substitution, which will transform our quartic equation into a quadratic equation – something we're much more familiar with.
When faced with solving equations like these, the initial instinct might be to try and factor it directly. While that's a valid approach, it can sometimes be challenging to spot the factors immediately, especially with higher-degree polynomials. That's where the beauty of substitution comes in. It allows us to simplify the equation, making it easier to manipulate and ultimately find the solutions. The key is recognizing patterns and choosing the right substitution to transform the equation into a more familiar form. In this case, the presence of both x^4 and x^2 terms suggests a quadratic-like structure, making substitution an ideal strategy.
The beauty of mathematics lies in its ability to transform complex problems into simpler ones. By recognizing patterns and applying appropriate techniques, we can unravel even the most daunting equations. So, let's embrace the power of substitution and see how it helps us conquer this quartic equation. Stay tuned as we delve into the substitution method and reveal the secrets to finding the real solutions of x^4 - 13x^2 + 36 = 0.
The Substitution Trick: Transforming Quartic to Quadratic
Alright, guys, let's get down to the nitty-gritty and apply the substitution method. This is where things get really interesting! Remember how we mentioned the equation has a special form? Well, notice that we have both x^4 and x^2 terms. This hints at a hidden quadratic structure. To reveal it, we'll make a clever substitution. Let's say:
y = x^2
Why this substitution? Because if y = x^2, then y^2 = (x2)2 = x^4. Suddenly, our quartic equation starts to look a lot more quadratic-like!
Now, let's rewrite our original equation x^4 - 13x^2 + 36 = 0 using this substitution. Replacing x^4 with y^2 and x^2 with y, we get:
y^2 - 13y + 36 = 0
Ta-da! We've successfully transformed our quartic equation into a quadratic equation in terms of 'y'. This is a huge win because we have well-established methods for solving quadratic equations, like factoring, completing the square, or using the quadratic formula.
The substitution method is a powerful technique in algebra because it allows us to simplify complex equations by introducing new variables. The key is to choose a substitution that transforms the equation into a more manageable form. In this case, recognizing the relationship between x^4 and x^2 was crucial in guiding our choice of substitution. By letting y = x^2, we effectively reduced the degree of the equation, making it easier to solve.
Think of it like this: we've taken a seemingly complicated puzzle and broken it down into smaller, more manageable pieces. The substitution is like a magic key that unlocks a simpler version of the problem. Now that we have a quadratic equation, we can employ our familiar tools to find the values of 'y'. But remember, we're not done yet! We need to go back and find the values of 'x' that satisfy the original equation. This is where the final step of the substitution method comes in – substituting back to find the original variable.
So, buckle up as we move on to solving the quadratic equation for 'y'. Once we have the values of 'y', we'll use our substitution y = x^2 to find the corresponding values of 'x'. The solution set for the original quartic equation is within our grasp, guys! Let's keep the momentum going.
Solving the Quadratic: Finding the Values of 'y'
Now that we've got our quadratic equation, y^2 - 13y + 36 = 0, it's time to crack it! There are several ways we can tackle this, but factoring is often the quickest and most elegant method if it's possible. So, let's see if we can factor this bad boy.
We're looking for two numbers that multiply to 36 (the constant term) and add up to -13 (the coefficient of the 'y' term). After a little thought, we can see that -4 and -9 fit the bill perfectly:
- (-4) * (-9) = 36
- (-4) + (-9) = -13
Awesome! This means we can factor our quadratic equation as follows:
(y - 4)(y - 9) = 0
Now, the zero-product property comes to our rescue. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if (y - 4)(y - 9) = 0, then either (y - 4) = 0 or (y - 9) = 0 (or both!).
Setting each factor equal to zero, we get two simple linear equations:
- y - 4 = 0 => y = 4
- y - 9 = 0 => y = 9
Fantastic! We've found the two solutions for 'y': y = 4 and y = 9. But remember, our ultimate goal is to find the values of 'x'. We're not done yet! We need to go back to our original substitution and use these values of 'y' to find the corresponding values of 'x'.
Factoring is a powerful tool for solving quadratic equations because it allows us to break down the equation into simpler factors. The zero-product property then provides a direct link between the factors and the solutions. However, if factoring isn't immediately obvious, we can always resort to other methods like the quadratic formula or completing the square. These methods are guaranteed to work for any quadratic equation, even those that are difficult to factor.
In this case, factoring provided a quick and efficient path to the solutions for 'y'. But the journey isn't over! We've only solved for the intermediate variable. The real magic happens when we substitute back and unveil the solutions for 'x'. So, let's keep the momentum going as we take the final step in solving our quartic equation.
The Final Step: Finding the Values of 'x'
Okay, folks, we're in the home stretch! We've found the values of 'y', which are y = 4 and y = 9. Now, we need to use our original substitution, y = x^2, to find the corresponding values of 'x'. This is where we'll finally uncover the solution set for our quartic equation.
Let's tackle each value of 'y' separately:
Case 1: y = 4
Substituting y = 4 into y = x^2, we get:
x^2 = 4
To solve for 'x', we take the square root of both sides. Remember, when taking the square root, we need to consider both the positive and negative roots:
x = ±√4
So, we get two solutions for this case:
- x = 2
- x = -2
Case 2: y = 9
Similarly, substituting y = 9 into y = x^2, we get:
x^2 = 9
Taking the square root of both sides, we again consider both positive and negative roots:
x = ±√9
This gives us two more solutions:
- x = 3
- x = -3
We've done it! We've found all the possible values of 'x' that satisfy the equation x^4 - 13x^2 + 36 = 0. We have four solutions in total: x = 2, x = -2, x = 3, and x = -3.
Remember, the substitution method is like a bridge that connects two different worlds – the world of quartic equations and the world of quadratic equations. By substituting and then substituting back, we can leverage our knowledge of quadratic equations to solve more complex problems. The key is to keep track of the original variables and to ensure that we substitute back to find the final solutions.
We've successfully navigated the intricacies of this quartic equation. Now, let's gather our results and present the solution set in a clear and concise manner.
The Grand Finale: The Solution Set
Alright, let's put it all together! We've diligently worked through the equation x^4 - 13x^2 + 36 = 0, using the power of substitution to transform it into a manageable quadratic equation. We solved for 'y' and then skillfully substituted back to find the values of 'x'.
Our efforts have revealed four distinct real number solutions:
- x = 2
- x = -2
- x = 3
- x = -3
Therefore, the solution set for the equation in the real number field is:
{2, -2, 3, -3}
This corresponds to option A) {2, -2, 3, -3} in the original problem.
So, we've not only found the solution set but also justified our answer by clearly explaining the method we used. We started by recognizing the quartic nature of the equation and the potential for substitution. We then strategically substituted y = x^2 to transform the equation into a quadratic form. We solved the quadratic equation by factoring, found the values of 'y', and finally, substituted back to obtain the values of 'x'.
Presenting the solution set in a clear and organized manner is crucial in mathematics. It allows others to easily understand our findings and verify our results. The solution set is a concise way to represent all the values that satisfy the original equation.
In conclusion, we've successfully unraveled the solution set of the quartic equation x^4 - 13x^2 + 36 = 0 using the substitution method. The solution set in the real number field is {2, -2, 3, -3}, which corresponds to option A. We've demonstrated the power of algebraic manipulation and the elegance of the substitution technique in solving complex equations. Great job, everyone! We conquered this mathematical challenge together.
