Solving Radical Equation What Is The Solution Of $\sqrt{-4x} = 100$

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Hey guys! Let's dive into solving this radical equation step by step. We've got βˆ’4x=100\sqrt{-4x} = 100 and we need to figure out what value of x makes this true. Radical equations can seem tricky, but with a clear approach, we can crack this one.

Understanding the Problem

Before we jump into solving, let's understand what we're dealing with. The equation βˆ’4x=100\sqrt{-4x} = 100 involves a square root, which means we're looking for a number that, when square-rooted, gives us 100. However, there's also a negative sign inside the square root, which adds a layer of complexity.

The key here is to remember that the square root of a number is only defined for non-negative values within the realm of real numbers. This means that the expression inside the square root, which is -4x in our case, must be greater than or equal to zero. If -4x is negative, then the square root will result in an imaginary number, which we typically don't deal with in basic algebra unless specifically asked for.

So, our initial consideration should be whether there's any real number solution at all. If -4x cannot be non-negative, then there will be no real solution to the equation.

Step-by-Step Solution

1. Square Both Sides

The first step to solving a radical equation is usually to get rid of the square root. We can do this by squaring both sides of the equation. Squaring both sides helps us eliminate the square root, making it easier to isolate x.

So, starting with our equation:

βˆ’4x=100\sqrt{-4x} = 100

Squaring both sides, we get:

(βˆ’4x)2=1002(\sqrt{-4x})^2 = 100^2

This simplifies to:

βˆ’4x=10000-4x = 10000

2. Isolate x

Now that we've eliminated the square root, we have a simple linear equation. To isolate x, we need to divide both sides of the equation by -4. This will give us the value of x.

βˆ’4x=10000-4x = 10000

Dividing both sides by -4:

x=10000βˆ’4x = \frac{10000}{-4}

Simplifying, we get:

x=βˆ’2500x = -2500

3. Check the Solution

An important step in solving radical equations is to check our solution. This is because squaring both sides of an equation can sometimes introduce extraneous solutionsβ€”solutions that satisfy the transformed equation but not the original one. We need to make sure that our solution, x = -2500, actually works in the original equation.

Let's plug x = -2500 back into the original equation:

βˆ’4x=100\sqrt{-4x} = 100

βˆ’4(βˆ’2500)=100\sqrt{-4(-2500)} = 100

10000=100\sqrt{10000} = 100

100=100100 = 100

Our solution checks out! When we substitute x = -2500 into the original equation, it holds true. This confirms that x = -2500 is indeed a valid solution.

Analyzing the Solution

Why Checking is Crucial

Checking the solution in radical equations is not just a formality; it’s a crucial step. Squaring both sides of an equation can introduce solutions that don't actually work in the original equation. These are called extraneous solutions. For instance, consider the equation x=βˆ’5\sqrt{x} = -5. If we square both sides, we get x=25x = 25. However, if we plug x=25x = 25 back into the original equation, we get 25=5\sqrt{25} = 5, not -5. Thus, x=25x = 25 is an extraneous solution.

The Role of the Negative Sign

The negative sign inside the square root in our equation, βˆ’4x=100\sqrt{-4x} = 100, might have initially raised a red flag. Remember, the expression inside the square root must be non-negative for the solution to be real. When we found x = -2500, we saw that βˆ’4x-4x becomes βˆ’4(βˆ’2500)=10000-4(-2500) = 10000, which is positive. This is why our solution works.

Implications of the Solution

The solution x = -2500 tells us that there is one real value of x that satisfies the equation βˆ’4x=100\sqrt{-4x} = 100. This value makes the expression inside the square root equal to 10000, which has a real square root of 100. This understanding is vital for solving similar equations and grasping the properties of radical expressions.

Common Mistakes to Avoid

Forgetting to Check Solutions

One of the biggest mistakes students make when solving radical equations is forgetting to check their solutions. As we've seen, checking is essential to avoid extraneous solutions. Always plug your solution back into the original equation to verify its validity.

Incorrectly Squaring Both Sides

Another common mistake is squaring both sides of the equation incorrectly. Make sure to square the entire side, not just individual terms. For example, if you have an equation like x+2=5\sqrt{x} + 2 = 5, you need to square (x+2)(\sqrt{x} + 2), not just x\sqrt{x} and 2 separately.

Ignoring the Domain of Square Roots

It’s also crucial to remember that the expression inside a square root must be non-negative for real solutions. Ignoring this can lead to incorrect solutions or overlooking the possibility of no real solutions.

Misunderstanding Negative Signs

Negative signs, like the one in βˆ’4x\sqrt{-4x}, can be confusing. Always consider how the negative sign affects the expression and the possible solutions. In our case, the negative sign allowed for a negative value of x, but it's crucial to ensure that the overall expression inside the square root remains non-negative.

Alternative Methods and Insights

Graphical Approach

Another way to approach this problem is graphically. We can think of the equation βˆ’4x=100\sqrt{-4x} = 100 as finding the intersection of two functions: y=βˆ’4xy = \sqrt{-4x} and y=100y = 100. The x-coordinate of the intersection point will be the solution to the equation.

The graph of y=βˆ’4xy = \sqrt{-4x} is a reflection of the standard square root function y=xy = \sqrt{x} across the y-axis, compressed horizontally by a factor of 4. The graph of y=100y = 100 is a horizontal line. By plotting these two functions, we would see that they intersect at the point (βˆ’2500,100)(-2500, 100), confirming our algebraic solution.

Domain Considerations

Before even solving the equation, we can consider the domain of the function y=βˆ’4xy = \sqrt{-4x}. For the square root to be real, we need βˆ’4xβ‰₯0-4x \geq 0. Dividing both sides by -4 (and flipping the inequality sign because we're dividing by a negative number), we get x≀0x \leq 0. This tells us that any solution for x must be non-positive. Our solution, x = -2500, fits within this domain.

Complex Numbers

If we were considering complex numbers, the situation would be different. The square root of a negative number is defined in terms of the imaginary unit i, where i=βˆ’1i = \sqrt{-1}. However, in the context of this problem, we are looking for real solutions, so we focus on the non-negativity of the expression inside the square root.

Real-World Applications

Radical equations might seem abstract, but they have real-world applications in various fields. Here are a few examples:

Physics

In physics, radical equations often appear when dealing with motion and energy. For example, the velocity of an object falling under gravity can be described using a radical equation. Similarly, the period of a pendulum can be calculated using a formula involving a square root.

Engineering

Engineers use radical equations in various calculations, such as determining the strength of materials or designing structures. The stress and strain on a material under load can sometimes be related through radical equations.

Finance

In finance, radical equations can be used to calculate growth rates or investment returns. For example, the compound interest formula involves radicals, and solving for the interest rate might require solving a radical equation.

Computer Graphics

Radical equations are used in computer graphics for rendering and transformations. Calculations involving distances and angles often involve square roots, which can lead to radical equations.

Conclusion

So, the solution to the equation βˆ’4x=100\sqrt{-4x} = 100 is x = -2500. Remember, guys, to always check your solutions in radical equations to avoid extraneous results. Understanding the underlying principles and common pitfalls will make you a pro at solving these types of problems. Keep practicing, and you’ll master them in no time! We've walked through the steps, checked our answer, and discussed why this solution makes sense. Keep up the great work, and you'll become a master at solving radical equations!

Final Answer: A. x=βˆ’2500x = -2500