Problem 55 - Algebra and Trigonometry

Question

\(47-70\) The given equation involves a power of the variable. Find all real solutions of the equation. $$ (x+2)^{2}=4 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 0\) and \(x = -4\).

Step 1: Understand the Equation

We are given the equation \((x+2)^2 = 4\). This is a quadratic equation where the square of a binomial is equal to a constant.

Step 2: Apply the Square Root Property

To solve the equation \((x+2)^2 = 4\), we apply the square root property. This gives two solutions: \(x+2 = \pm \sqrt{4}\), which simplifies to \(x+2 = 2\) or \(x+2 = -2\).

Step 3: Solve for x

Solve the equations from the previous step: \(x+2 = 2\) results in \(x = 0\) and \(x+2 = -2\) gives \(x = -4\).

Step 4: Verify the Solutions

Substitute \(x = 0\) and \(x = -4\) back into the original equation to verify: \((0+2)^2 = 4\) and \((-4+2)^2 = 4\), thus both solutions satisfy the equation.

Key Concepts

Real Solutions

In mathematics, finding real solutions for an equation involves identifying values for the variable that satisfy the equation within the realm of real numbers. Real numbers include all the numbers on the number line, encompassing both rational and irrational numbers.
The original equation given is \((x+2)^{2}=4\). Since this equation involves a power of 2, we are seeking solutions where the left side of the equation equals 4 in the real number system.
To determine how many real solutions exist, solve the equation using algebraic techniques such as factoring, completing the square, or applying properties like the square root property.

• A solution is considered "real" if it doesn't involve any imaginary numbers, represented typically with an "i".
• In this context, the square of a binomial \((x+2)\) results in real numbers \(2\) and \(-2\) upon solving, indicating the presence of two real solutions.

Square Root Property

The Square Root Property is a powerful tool for solving quadratic equations that involve squares. This property states that if \(a^{2} = b\), then \(a\) can be \(\pm \sqrt{b}\). This means we must consider both the positive and negative roots.
Applying this property to \((x+2)^{2} = 4\):

• First, we take the square root of both sides, resulting in \(x+2 = \pm \sqrt{4}\).
• Since the square root of 4 is 2, we derive \(x+2 = 2\) and \(x+2 = -2\).

By isolating \(x\) in each equation, we uncover the solutions \(x=0\) and \(x=-4\).
This procedure highlights the necessity of considering both positive and negative roots when using the square root property, which is a common step in solving quadratic equations.

Verification of Solutions

Verification is the crucial last step in solving any equation. It ensures that the computed solutions actually satisfy the original equation, confirming their validity. Verification involves substituting the solutions back into the original equation.
For the solutions \(x=0\) and \(x=-4\):

• Substituting \(x=0\) into \((x+2)^{2}=4\), we have \((0+2)^{2} = 4\). This simplifies to \(4 = 4\), confirming that \(x=0\) is indeed a solution.
• Substituting \(x=-4\) into \((x+2)^{2}=4\), we have \((-4+2)^{2} = 4\). This also simplifies to \(4 = 4\), showing \(x=-4\) is a valid solution as well.

Correct verification is essential because it helps avoid errors from mathematical steps like sign flips or arithmetic slips. It's a simple yet effective way to corroborate that the obtained solutions are, without doubt, the correct ones.