Solving For X: Where Does √(1-x²) = 2x-1?

Are you wrestling with the challenge of finding where two seemingly different equations intersect? Let's break down the process of solving for the approximate x-value where the equations y = √(1 - x²) and y = 2x - 1 are equal. This problem combines algebra and a bit of geometric intuition, so buckle up and let's dive in!

Understanding the Equations

Before we jump into solving, let's quickly understand what these equations represent. The first equation, y = √(1 - x²), might look a bit intimidating, but it's actually the upper half of a circle with a radius of 1, centered at the origin (0, 0). Think about it: if you square both sides, you get y² = 1 - x², which can be rearranged to x² + y² = 1. This is the standard equation of a unit circle. Since we only have the square root, we're only considering the positive y values, hence the upper half.

The second equation, y = 2x - 1, is much simpler – it's a straight line with a slope of 2 and a y-intercept of -1. Our goal is to find the x-coordinate where this line intersects the upper half of the circle. This involves a bit of algebraic manipulation and understanding of the properties of both equations. The key to solving this problem lies in combining these equations effectively. By setting them equal to each other, we can eliminate y and create a single equation in terms of x. This allows us to use algebraic techniques, such as squaring both sides, to simplify the equation and isolate x. However, it's crucial to be mindful of potential extraneous solutions that may arise from squaring, which we'll address later in the process. Furthermore, understanding the graphical representation of the equations – a semicircle and a line – can provide valuable intuition about the number and approximate location of the solutions. This visual perspective can help us verify the reasonableness of our algebraic solutions and identify any potential errors in our calculations. In essence, solving this problem requires a blend of algebraic skill and geometric insight to navigate the complexities of the equations and arrive at the correct answer.

Setting the Equations Equal

To find the point(s) where these two equations are equal, we can set them equal to each other: √(1 - x²) = 2x - 1. This is the crucial first step in solving for the x-value where the semicircle and the line intersect. By equating the two expressions for y, we effectively create a single equation in one variable, x, which we can then manipulate algebraically to find its value. This approach leverages the fundamental concept that at the point of intersection, the y-values of both equations must be the same. Therefore, setting the right-hand sides of the equations equal to each other allows us to find the corresponding x-value(s) at that point. However, this step is just the beginning of the solution process. The resulting equation, √(1 - x²) = 2x - 1, involves a square root, which necessitates further algebraic manipulation to isolate x. This typically involves squaring both sides of the equation, a process that can introduce extraneous solutions. Therefore, it's essential to check the solutions obtained at the end to ensure they satisfy the original equation. Moreover, understanding the potential for extraneous solutions underscores the importance of a thorough and careful approach to solving equations involving radicals. By setting the equations equal, we initiate the process of finding the intersection point(s), but we must remain vigilant about the algebraic steps that follow to ensure the accuracy and validity of our final answer. This initial step, while seemingly straightforward, sets the stage for a more complex algebraic journey that demands precision and attention to detail.

Squaring Both Sides

To get rid of the square root, we'll square both sides of the equation: (√(1 - x²))² = (2x - 1)². This step is a pivotal moment in the solution process as it eliminates the square root, allowing us to work with a more manageable algebraic expression. Squaring both sides of the equation effectively transforms it from a radical equation into a polynomial equation, which is generally easier to solve. However, this transformation comes with a critical caveat: it can introduce extraneous solutions. Extraneous solutions are values of x that satisfy the squared equation but do not satisfy the original equation. This occurs because the squaring operation can obscure the original domain restrictions or introduce new solutions that were not present before. For example, if we have an equation like √x = -2, squaring both sides gives x = 4, but 4 is not a valid solution to the original equation because the square root of a number cannot be negative. In our case, squaring both sides of √(1 - x²) = 2x - 1 will lead to a quadratic equation, which may have two solutions. It's imperative that we check both of these solutions in the original equation to determine if they are valid or extraneous. Therefore, while squaring both sides is a necessary step to simplify the equation, it also adds a layer of complexity to the solution process. We must proceed with caution and remember to verify our final answers to avoid accepting solutions that do not truly satisfy the initial equation. This careful approach is crucial for ensuring the accuracy and integrity of our solution.

Expanding the right side, we get: 1 - x² = 4x² - 4x + 1. The expansion of (2x - 1)² into 4x² - 4x + 1 is a direct application of the binomial expansion formula or simply multiplying the expression by itself. This step is crucial in transforming the equation into a more recognizable form, specifically a quadratic equation, which we can then solve using standard techniques. The accuracy of this expansion is paramount, as any error here will propagate through the rest of the solution and lead to an incorrect final answer. Each term must be multiplied correctly, paying careful attention to signs and coefficients. This meticulous approach ensures that the resulting quadratic equation accurately represents the relationship between x and y after the square root has been eliminated. Furthermore, the expanded form allows us to collect like terms and rearrange the equation into the standard quadratic form, which is essential for applying methods such as factoring, completing the square, or using the quadratic formula. Therefore, this step is not just a mechanical expansion; it's a crucial transformation that sets the stage for the subsequent steps in solving for x. By expanding the expression correctly, we pave the way for a clear and accurate path to the solution.

Simplifying to a Quadratic Equation

Now, let's simplify and rearrange the equation into a standard quadratic form (ax² + bx + c = 0): 0 = 5x² - 4x. This step involves combining like terms and rearranging them to achieve the standard quadratic form, which is a necessary precursor to solving for x. By moving all terms to one side of the equation, we set the stage for applying various solution methods, such as factoring, completing the square, or using the quadratic formula. The goal is to create an equation where one side is equal to zero, allowing us to utilize the zero-product property or other techniques specific to quadratic equations. The accuracy of this simplification process is crucial, as any errors in combining or rearranging terms will lead to an incorrect quadratic equation and, consequently, incorrect solutions for x. It's essential to pay close attention to signs and coefficients while performing these operations. Moreover, recognizing the standard quadratic form is a fundamental skill in algebra, as it allows us to readily identify the coefficients a, b, and c, which are necessary for applying the quadratic formula or other solution methods. Therefore, this step is not merely a mechanical rearrangement of terms; it's a critical transformation that enables us to leverage the well-established tools and techniques for solving quadratic equations, ultimately bringing us closer to finding the desired x-value(s).

Solving the Quadratic Equation

We can factor out an x from the equation: 0 = x(5x - 4). This factorization step is a powerful technique for solving quadratic equations, particularly when the equation can be easily factored. By identifying and extracting the common factor x, we transform the quadratic equation into a product of two factors, one of which is simply x and the other is a linear expression (5x - 4). This factorization is based on the distributive property in reverse and allows us to apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property is the key to finding the solutions for x, as it allows us to set each factor equal to zero and solve the resulting simpler equations. The ease of this factorization underscores the importance of recognizing common factors in algebraic expressions, as it can significantly simplify the solution process. However, not all quadratic equations can be easily factored, and in such cases, other methods like completing the square or using the quadratic formula may be necessary. In this particular instance, the factorization method provides a direct and efficient path to finding the potential solutions for x. By successfully factoring the equation, we set the stage for applying the zero-product property and quickly determining the values of x that satisfy the equation.

Setting each factor to zero gives us two potential solutions: x = 0 or 5x - 4 = 0. Solving the second equation, we get x = 4/5 = 0.8. This step is a direct application of the zero-product property, which is a fundamental principle in algebra for solving factored equations. The zero-product property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. In our case, the factored equation 0 = x(5x - 4) implies that either x = 0 or 5x - 4 = 0. By setting each factor equal to zero, we create two separate equations that are much simpler to solve than the original quadratic equation. The equation x = 0 directly gives us one potential solution, while the equation 5x - 4 = 0 requires a simple algebraic step to isolate x. Adding 4 to both sides and then dividing by 5 yields x = 4/5, which is equivalent to 0.8. These two potential solutions, x = 0 and x = 0.8, represent the x-values where the product of the factors is zero, and thus, are the candidates for the solutions of the original quadratic equation. However, it's crucial to remember that we squared the original equation earlier, which could have introduced extraneous solutions. Therefore, we must verify these potential solutions in the original equation to ensure they are valid.

Checking for Extraneous Solutions

Remember, squaring both sides can introduce extraneous solutions, so we need to check our answers in the original equation, y = √(1 - x²) and y = 2x - 1.

Let's check x = 0:

• y = √(1 - 0²) = 1
• y = 2(0) - 1 = -1

The y values don't match, so x = 0 is an extraneous solution. This check is a crucial step in the solution process, particularly when dealing with equations involving square roots or other operations that can introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (in this case, the squared equation) but do not satisfy the original equation. Failing to identify and discard these solutions can lead to incorrect answers. In the case of x = 0, substituting it into the first original equation, y = √(1 - x²), yields y = √(1 - 0²) = √1 = 1. However, substituting x = 0 into the second original equation, y = 2x - 1, gives y = 2(0) - 1 = -1. Since the y-values obtained from the two equations do not match, x = 0 does not represent a point of intersection between the two curves and is therefore an extraneous solution. This discrepancy highlights the importance of verifying potential solutions in the original equations to ensure they are valid. By performing this check, we can confidently eliminate extraneous solutions and focus on the true solutions of the problem. This careful approach is essential for maintaining the accuracy and integrity of our solution process.

Now let's check x = 0.8:

• y = √(1 - (0.8)²) = √(1 - 0.64) = √0.36 = 0.6
• y = 2(0.8) - 1 = 1.6 - 1 = 0.6

The y values match, so x = 0.8 is a valid solution. This verification step is the final confirmation that our solution is correct and satisfies the original problem. By substituting x = 0.8 into both of the original equations, we can determine the corresponding y-values and check if they are equal. This equality is the definitive proof that the point (x, y) lies on both the semicircle and the line, and therefore represents a valid intersection point. Substituting x = 0.8 into y = √(1 - x²) yields y = √(1 - (0.8)²) = √(1 - 0.64) = √0.36 = 0.6. Similarly, substituting x = 0.8 into y = 2x - 1 gives y = 2(0.8) - 1 = 1.6 - 1 = 0.6. Since both equations produce the same y-value of 0.6, we can confidently conclude that x = 0.8 is a valid solution. This consistency in the y-values confirms that the point (0.8, 0.6) is indeed a point of intersection between the semicircle and the line. Therefore, this verification step not only validates our algebraic solution but also provides a geometric interpretation of the result. By confirming the solution, we complete the problem-solving process and gain a deeper understanding of the relationship between the two equations.

The Answer

The approximate x-value where the two equations are equal is 0.8. So the answer is B. This final conclusion is the culmination of all the steps we've taken, from setting the equations equal to each other to verifying the solutions. The process has involved algebraic manipulation, including squaring both sides of the equation, simplifying, factoring, and solving for x. It has also required a critical awareness of extraneous solutions and the importance of checking potential answers in the original equations. The result, x = 0.8, is the x-coordinate of the point where the semicircle and the line intersect, representing the solution to the original problem. This answer not only satisfies the equations but also has a geometric significance, representing a specific point on the coordinate plane. The journey to this solution has underscored the importance of precision, attention to detail, and a thorough understanding of algebraic principles. By carefully executing each step and verifying our results, we can confidently arrive at the correct answer and gain a deeper appreciation for the interplay between algebra and geometry. The final answer of x = 0.8 is not just a number; it's the result of a logical and systematic problem-solving process.

In conclusion, solving for the intersection of equations often involves a multi-step process. In this case, we equated the two equations, manipulated the result to eliminate the square root, solved the resulting quadratic, and crucially, checked for extraneous solutions. This careful approach ensures we arrive at the correct answer. For further exploration of solving equations and mathematical concepts, check out Khan Academy's algebra section.