Determine The Solution For X2 36 12x

Solving the Quadratic Equation: x² + 12x - 36 = 0

Determining the solution for the quadratic equation x² + 12x - 36 = 0 involves finding the values of 'x' that satisfy the equation. Quadratic equations, characterized by their highest power of x being 2, can be solved using several methods. We'll explore three common and effective techniques: factoring, the quadratic formula, and completing the square. Each method offers a unique approach, allowing for flexibility based on the specific equation and personal preference. Understanding these methods will empower you to tackle a wide range of quadratic problems.

Method 1: Factoring the Quadratic Expression

Factoring involves expressing the quadratic expression as a product of two linear expressions. This method is often the quickest and easiest if the quadratic expression is easily factorable. Let's see if we can factor x² + 12x - 36.

We're looking for two numbers that add up to 12 (the coefficient of x) and multiply to -36 (the constant term). Let's consider the factors of -36:

• 1 and -36
• 2 and -18
• 3 and -12
• 4 and -9
• 6 and -6

Notice that none of these pairs add up to 12. This means that the quadratic expression x² + 12x - 36 cannot be factored easily using integers. Therefore, factoring is not the most efficient method in this case. We'll need to resort to other techniques.

Method 2: The Quadratic Formula

The quadratic formula is a powerful tool that works for all quadratic equations, regardless of whether they are easily factorable or not. It provides a direct way to calculate the solutions (also known as roots or zeros) of the equation ax² + bx + c = 0. The formula is:

x = [-b ± √(b² - 4ac)] / 2a

In our equation, x² + 12x - 36 = 0, we have a = 1, b = 12, and c = -36. Let's substitute these values into the quadratic formula:

x = [-12 ± √(12² - 4 * 1 * -36)] / (2 * 1)

x = [-12 ± √(144 + 144)] / 2

x = [-12 ± √288] / 2

x = [-12 ± √(144 * 2)] / 2

x = [-12 ± 12√2] / 2

Now we can simplify:

x = -6 ± 6√2

Therefore, the two solutions for x are:

• x₁ = -6 + 6√2
• x₂ = -6 - 6√2

These are the exact solutions. You can approximate these values using a calculator:

• x₁ ≈ 2.485
• x₂ ≈ -14.485

Method 3: Completing the Square

Completing the square is another powerful technique used to solve quadratic equations. This method involves manipulating the equation to create a perfect square trinomial, which can then be easily factored.

1. Move the constant term to the right side:

   x² + 12x = 36

2. Take half of the coefficient of x (which is 12), square it (6² = 36), and add it to both sides:

   x² + 12x + 36 = 36 + 36

3. Factor the left side as a perfect square:

   (x + 6)² = 72

4. Take the square root of both sides:

   x + 6 = ±√72

5. Simplify the square root:

   x + 6 = ±√(36 * 2) = ±6√2

6. Solve for x:

   x = -6 ± 6√2

This gives us the same solutions as the quadratic formula:

• x₁ = -6 + 6√2
• x₂ = -6 - 6√2

Understanding the Solutions and Their Significance

The solutions x₁ = -6 + 6√2 and x₂ = -6 - 6√2 represent the x-intercepts (points where the graph of the quadratic equation intersects the x-axis) of the parabola represented by the equation y = x² + 12x - 36. These points indicate where the value of the quadratic expression equals zero.

The fact that we have two distinct real solutions indicates that the parabola intersects the x-axis at two different points. If the discriminant (b² - 4ac) in the quadratic formula were negative, we would have complex conjugate solutions, meaning the parabola would not intersect the x-axis. If the discriminant were zero, we would have one repeated real solution (the parabola would touch the x-axis at only one point).

Verifying the Solutions

It's always a good practice to verify your solutions by substituting them back into the original equation:

For x₁ = -6 + 6√2:

(-6 + 6√2)² + 12(-6 + 6√2) - 36 =

36 - 72√2 + 72 - 72 + 72√2 - 36 = 0

For x₂ = -6 - 6√2:

(-6 - 6√2)² + 12(-6 - 6√2) - 36 =

36 + 72√2 + 72 - 72 - 72√2 - 36 = 0

Both solutions satisfy the original equation, confirming their correctness.

Applications of Quadratic Equations

Quadratic equations have numerous applications across various fields, including:

• Physics: Calculating projectile motion, determining the path of a ball thrown in the air.
• Engineering: Designing structures, analyzing the strength of materials.
• Economics: Modeling supply and demand, calculating profit maximization.
• Computer graphics: Creating curved lines and shapes.

Understanding how to solve quadratic equations is a fundamental skill in mathematics and has widespread practical applications. The methods discussed – factoring, the quadratic formula, and completing the square – provide a comprehensive toolkit for tackling these equations effectively. Remember to choose the method that best suits the specific equation and your comfort level. Practicing these methods will build your confidence and proficiency in solving quadratic equations.