Solving 2x = √(5x² + 4x - 12): A Step-by-Step Guide

Hey Plastik Magazine readers! Today, we're diving into a fun mathematical problem that involves solving for x in the equation 2x = √(5x² + 4x - 12). This type of problem might seem intimidating at first, but don't worry, we'll break it down step by step. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the question is asking. We're given an equation where 2x is equal to the square root of a quadratic expression (5x² + 4x - 12). Our goal is to find all real numbers x that satisfy this equation. Remember, a real number is any number that can be found on the number line – think integers, fractions, decimals, and even irrational numbers like √2 or π.

This equation involves a square root, which means we need to be extra careful. Squaring both sides of an equation is a common technique for dealing with square roots, but it can sometimes introduce extraneous solutions. Extraneous solutions are values that we find during the solving process that don't actually work when we plug them back into the original equation. So, it's super important to check our answers at the end!

To solve this, we'll need to utilize our knowledge of algebraic manipulation, quadratic equations, and a bit of careful thinking. Let's break down the key concepts we'll be using:

• Squaring both sides: If a = b, then a² = b². This will help us get rid of the square root.
• Quadratic equations: These are equations of the form ax² + bx + c = 0. We can solve them by factoring, using the quadratic formula, or completing the square.
• Checking for extraneous solutions: This is crucial! After we find potential solutions, we need to plug them back into the original equation to make sure they actually work.

With these concepts in mind, let's roll up our sleeves and tackle this equation!

Step 1: Squaring Both Sides

The first step in solving this equation is to eliminate the square root. The easiest way to do this is by squaring both sides of the equation. Remember, whatever we do to one side of the equation, we must do to the other to maintain the balance. So, let's square both sides:

(2x)² = (√(5x² + 4x - 12))²

When we square 2x, we get 4x². When we square the square root, it simply cancels out, leaving us with the expression inside the square root. This gives us:

4x² = 5x² + 4x - 12

Great! We've successfully eliminated the square root and now have a quadratic equation to work with. This is a significant step forward. By squaring both sides, we've transformed the equation into a more manageable form that we can solve using standard algebraic techniques. However, as we mentioned earlier, this step is where extraneous solutions can creep in, so we'll need to be extra vigilant about checking our answers later.

Now that we have a quadratic equation, our next step will be to rearrange it into the standard form (ax² + bx + c = 0) so we can easily solve it. This will involve moving all the terms to one side of the equation, which we'll do in the next step. Stay with me, guys, we're making progress!

Step 2: Rearranging the Equation

Now that we've squared both sides, we have the equation 4x² = 5x² + 4x - 12. To solve this quadratic equation, we need to get it into the standard form: ax² + bx + c = 0. This means we need to move all the terms to one side of the equation, leaving zero on the other side.

Let's subtract 4x² from both sides of the equation:

4x² - 4x² = 5x² + 4x - 12 - 4x²

This simplifies to:

0 = x² + 4x - 12

Excellent! We've successfully rearranged the equation into the standard quadratic form. Now we have a clear quadratic expression (x² + 4x - 12) set equal to zero. This is a crucial step because it allows us to use various methods to find the solutions for x, such as factoring, using the quadratic formula, or completing the square.

By rearranging the equation, we've set the stage for solving it. The next step is to choose a method for solving the quadratic equation. In this case, factoring might be the easiest approach, but we could also use the quadratic formula. We'll explore the factoring method in the next step. Keep up the great work!

Step 3: Solving the Quadratic Equation by Factoring

We've arrived at the quadratic equation x² + 4x - 12 = 0. One way to solve this is by factoring. Factoring involves breaking down the quadratic expression into two binomials (expressions with two terms) that multiply together to give us the original expression. In other words, we're looking for two expressions in the form (x + p)(x + q) that, when multiplied, equal x² + 4x - 12.

To factor this quadratic, we need to find two numbers, p and q, that satisfy two conditions:

1. Their product (p * q*) is equal to the constant term, which is -12.
2. Their sum (p + q) is equal to the coefficient of the x term, which is 4.

Let's think about the factors of -12. We have pairs like (1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), and (-3, 4). Which of these pairs adds up to 4? The pair (-2, 6) works perfectly! -2 multiplied by 6 equals -12, and -2 plus 6 equals 4.

So, we can factor the quadratic equation as follows:

x² + 4x - 12 = (x - 2)(x + 6) = 0

Now we have the factored form of the equation. This is super useful because it allows us to use the zero-product property. The zero-product 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 a * b = 0, then either a = 0 or b = 0 (or both).

Applying the zero-product property to our factored equation, we get two possible equations:

x - 2 = 0 or x + 6 = 0

Now we can easily solve for x in each equation.

Step 4: Finding Potential Solutions

From the factored equation (x - 2)(x + 6) = 0, we obtained two possible equations: x - 2 = 0 and x + 6 = 0. Now, let's solve each of these equations to find potential values for x.

For the equation x - 2 = 0, we add 2 to both sides:

x - 2 + 2 = 0 + 2

This gives us:

x = 2

So, one potential solution is x = 2.

Now, let's solve the second equation, x + 6 = 0. We subtract 6 from both sides:

x + 6 - 6 = 0 - 6

This gives us:

x = -6

So, another potential solution is x = -6.

We've found two potential solutions: x = 2 and x = -6. However, remember that we squared both sides of the equation earlier, which could have introduced extraneous solutions. Therefore, it's absolutely crucial to check these potential solutions in the original equation to see if they actually work.

The next step is to plug each of these values back into the original equation, 2x = √(5x² + 4x - 12), and see if they make the equation true. This is the final and most important step in ensuring we have the correct solutions. Let's move on to checking our answers!

Step 5: Checking for Extraneous Solutions

We've found two potential solutions for x: 2 and -6. Now, we need to check if these solutions actually satisfy the original equation, 2x = √(5x² + 4x - 12). This step is crucial to eliminate any extraneous solutions that may have been introduced when we squared both sides of the equation.

Let's start by checking x = 2. Plug it into the original equation:

2(2) = √(5(2)² + 4(2) - 12)

4 = √(5(4) + 8 - 12)

4 = √(20 + 8 - 12)

4 = √16

4 = 4

This equation is true! So, x = 2 is a valid solution.

Now, let's check x = -6. Plug it into the original equation:

2(-6) = √(5(-6)² + 4(-6) - 12)

-12 = √(5(36) - 24 - 12)

-12 = √(180 - 24 - 12)

-12 = √144

-12 = 12

This equation is not true! -12 does not equal 12. Therefore, x = -6 is an extraneous solution and must be discarded.

By checking our solutions, we've confirmed that x = 2 is a valid solution, but x = -6 is not. This highlights the importance of checking for extraneous solutions when dealing with equations involving square roots.

Final Answer

After carefully solving the equation 2x = √(5x² + 4x - 12) and checking for extraneous solutions, we've found that the only valid solution is x = 2.

So, the answer is:

x = 2

Great job, guys! We successfully solved the equation step by step, remembering to check for extraneous solutions along the way. This type of problem might seem tricky at first, but by breaking it down into smaller steps and understanding the key concepts, we can conquer any mathematical challenge. Keep practicing, and you'll become even more confident in your problem-solving abilities!