Solving X^2 = -11x + 4: Find The Roots

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Quadratic equations are a fundamental part of algebra, and understanding how to solve them is a crucial skill. Today, we're going to dive deep into solving a specific quadratic equation: $x^2 = -11x + 4$. This might look a little intimidating at first, but by the end of this article, you'll be well-equipped to find its solutions. We'll break down the process step-by-step, explaining the concepts involved and why each step is important. Our goal is to not only find the answers but also to build your confidence in tackling similar problems. We'll cover everything from rearranging the equation into standard form to applying the quadratic formula, ensuring you have a clear and comprehensive understanding. So, let's get started on this mathematical journey to unravel the solutions of $x^2 = -11x + 4$!

Understanding Quadratic Equations and Standard Form

Before we can solve our specific equation, $x^2 = -11x + 4$, it's essential to understand what a quadratic equation is and why we often rewrite it in a standard form. A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term that is squared. The general form of a quadratic equation is typically written as ax2+bx+c=0ax^2 + bx + c = 0, where 'a', 'b', and 'c' are coefficients (constants), and 'x' is the variable we are trying to solve for. The key characteristic is the x2x^2 term; without it, it wouldn't be quadratic. Understanding this standard form, ax2+bx+c=0ax^2 + bx + c = 0, is vital because most methods for solving quadratic equations rely on having the equation arranged this way. It allows us to easily identify the values of 'a', 'b', and 'c', which are crucial for applying formulas like the quadratic formula. In our case, $x^2 = -11x + 4$, we can see that it's not yet in the standard form ax2+bx+c=0ax^2 + bx + c = 0. The terms are spread across the equals sign. To get it into the standard form, we need to move all terms to one side of the equation, leaving zero on the other side. This involves basic algebraic manipulation. We can achieve this by adding 11x11x to both sides and subtracting 44 from both sides. Let's do that: x2+11x−4=0x^2 + 11x - 4 = 0. Now, this equation is in the standard quadratic form. By comparing x2+11x−4=0x^2 + 11x - 4 = 0 to ax2+bx+c=0ax^2 + bx + c = 0, we can clearly identify our coefficients: a=1a = 1 (since there's an implicit '1' multiplying x2x^2), b=11b = 11, and c=−4c = -4. Recognizing these coefficients is the first critical step in preparing to solve the equation using various algebraic techniques. This reorganization might seem simple, but it's the foundation upon which all subsequent solution methods are built. Without this standardized format, applying formulas or factoring techniques becomes significantly more challenging, if not impossible. It’s like preparing your ingredients before you start cooking; you need everything in place before you can begin the main task. So, whenever you encounter a quadratic equation, always aim to bring it to the ax2+bx+c=0ax^2 + bx + c = 0 form first.

Applying the Quadratic Formula: A Powerful Tool

Once we have our quadratic equation in standard form, ax2+bx+c=0ax^2 + bx + c = 0, we can employ a powerful and universally applicable tool: the quadratic formula. This formula provides the solutions for 'x' directly, regardless of whether the equation can be easily factored or not. The quadratic formula is derived from the process of completing the square on the general form of the quadratic equation and is given by: $x = rac-b eq eta imes ext{sqrt}(b^2 - 4ac)}{2a}$ . This formula might look a bit complex with its square roots and fractions, but it's incredibly straightforward to use once you understand its components. The symbols 'a', 'b', and 'c' are the coefficients we identified earlier when we put our equation into standard form. In our equation, x2+11x−4=0x^2 + 11x - 4 = 0, we found that a=1a = 1, b=11b = 11, and c=−4c = -4. Now, we simply substitute these values into the quadratic formula. Let's plug them in $x = rac{-(11) eq eta imes ext{sqrt((11)^2 - 4(1)(-4))}2(1)}$ . The next step is to simplify the expression under the square root, which is called the discriminant. The discriminant (b2−4acb^2 - 4ac) tells us about the nature of the solutions. In our case, the discriminant is (11)2−4(1)(−4)=121−(−16)=121+16=137(11)^2 - 4(1)(-4) = 121 - (-16) = 121 + 16 = 137. Since the discriminant (137) is a positive number, we know that our equation will have two distinct real solutions. Now, let's substitute this back into the formula $x = rac{-11 eq eta imes ext{sqrt(137)}2}$ . This gives us our two solutions. The '±' symbol means we have two possibilities one where we add the square root and one where we subtract it. So, the two solutions are: $x_1 = rac{-11 + ext{sqrt(137)}{2}$ and $x_2 = rac{-11 - ext{sqrt}(137)}{2}$ . These are the exact solutions to our equation. The quadratic formula is indispensable because it works for all quadratic equations, even those that are difficult or impossible to factor. It's a reliable method that guarantees you'll find the roots if they exist. Remember, the key is to accurately identify 'a', 'b', and 'c' and substitute them carefully into the formula. Precision in calculation is important here to avoid errors.

Calculating the Specific Solutions

Now that we have the general form of the solutions from the quadratic formula, let's focus on calculating the specific numerical values for our equation x2+11x−4=0x^2 + 11x - 4 = 0. We determined that a=1a = 1, b=11b = 11, and c=−4c = -4. Plugging these into the quadratic formula, we arrived at: $x = rac-11 eq eta imes ext{sqrt}(137)}{2}$ . The square root of 137 ($ ext{sqrt}(137)$) is an irrational number, meaning it cannot be expressed as a simple fraction and its decimal representation goes on forever without repeating. For many mathematical purposes, leaving the answer in terms of the square root is the most accurate and preferred method. This is known as the exact form of the solution. Therefore, the two solutions are precisely $x_1 = rac{-11 + ext{sqrt(137)}2}$ and $x_2 = rac{-11 - ext{sqrt}(137)}{2}$ . These expressions represent the exact mathematical answers. If an approximate decimal value is needed, we would use a calculator to find the approximate value of $ ext{sqrt}(137)$, which is about 11.705. Then we would substitute this approximate value $x_1 eq rac{-11 + 11.7052} = rac{0.705}{2} eq 0.3525$ and $x_2 eq rac{-11 - 11.705}{2} = rac{-22.705}{2} eq -11.3525$ . However, unless specifically asked for an approximation, the exact forms are usually preferred in mathematics because they are precise and free from rounding errors. These exact solutions are perfectly valid and complete answers to the problem. When you see options like A, B, or C in a multiple-choice question, you should look for the one that matches these exact forms. Comparing our derived solutions to the given options Option A: $ rac{-11- ext{sqrt(137)}2}, rac{-11+ ext{sqrt}(137)}{2}$ This perfectly matches our calculated solutions. Option B $ rac{-11- ext{sqrt(125)}2}, rac{-11+ ext{sqrt}(125)}{2}$ This option has $ ext{sqrt}(125)$ instead of $ ext{sqrt}(137)$. Since 125eq137125 eq 137, this option is incorrect. Option C $ rac{11- ext{sqrt(137)}{2}$ This option only provides one solution and also has 1111 instead of −11-11 in the numerator for the first part. It is also incorrect. Therefore, based on our step-by-step calculation using the quadratic formula, option A correctly represents the solutions to the equation $x^2 = -11x + 4$. This process highlights the importance of accurate substitution and calculation when using the quadratic formula.

Conclusion: Mastering Quadratic Solutions

We've successfully navigated the process of solving the quadratic equation $x^2 = -11x + 4$. By first rearranging the equation into its standard form, x2+11x−4=0x^2 + 11x - 4 = 0, we identified the coefficients a=1a=1, b=11b=11, and c=−4c=-4. Then, we applied the robust quadratic formula, $x = rac{-b eq ext{sqrt}(b^2 - 4ac)}{2a}$, substituting these values to find the solutions. The calculation yielded the discriminant b2−4ac=137b^2 - 4ac = 137, confirming two distinct real roots. Ultimately, the exact solutions were found to be $x = rac{-11 eq ext{sqrt}(137)}{2}$. This means our two solutions are $ rac{-11 + ext{sqrt}(137)}{2}$ and $ rac{-11 - ext{sqrt}(137)}{2}$. Recognizing these as the correct answers aligns perfectly with Option A from the provided choices. The journey of solving quadratic equations, from understanding their standard form to applying the quadratic formula, is a fundamental skill in mathematics. It empowers you to tackle a wide range of algebraic problems with confidence. Remember, practice is key! The more equations you solve, the more comfortable and proficient you will become. Don't be afraid to break down complex problems into smaller, manageable steps, just as we did today. Each step, from standardizing the equation to carefully plugging values into the formula, builds towards the final solution. Keep exploring, keep practicing, and you'll master quadratic equations in no time.

For further exploration and practice with quadratic equations, you can visit Khan Academy's comprehensive resources on algebra. They offer detailed explanations, practice exercises, and video tutorials that can further solidify your understanding of these important mathematical concepts. You can also find valuable information on Wolfram MathWorld, a fantastic resource for advanced mathematical topics and definitions.