Problem 57 - Algebra for College Students

Question

Find all real number solutions for each equation. $$x^{2}-25=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 5 \) and \( x = -5 \).

Step 1: Understand the Equation

The given equation is \( x^2 - 25 = 0 \). It's a quadratic equation in the form of \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 0 \), and \( c = -25 \). Our task is to find the values of \( x \) that satisfy this equation.

Step 2: Move Constant to Other Side

Add 25 to both sides of the equation to isolate the term with \( x \). This gives us \( x^2 = 25 \).

Step 3: Solve for x Using Square Roots

To solve \( x^2 = 25 \), take the square root of both sides of the equation. Remember that there are two possible solutions when taking the square root of a number, so \( x = \pm \sqrt{25} \).

Step 4: Evaluate the Square Roots

Calculate the value of the square root of 25. We have \( \sqrt{25} = 5 \). Thus, the solutions for \( x \) are \( x = 5 \) and \( x = -5 \).

Key Concepts

Real Number Solutions

When tackling quadratic equations like \(x^{2}-25=0\), we aim to find "real number solutions." These solutions are the values of \(x\) that make the equation true using real numbers without involving any imaginary components. In the case of our problem, \(x^2 = 25\) is modified from the original equation and has two potential real solutions.

Real numbers include all rational numbers (like 3, 1/2, or -5) and irrational numbers (like \(\sqrt{2}\) or \(\pi\)). To solve the quadratic equation, we consider all possible solutions within this set of numbers. For \(x^2 = 25\), both 5 and -5 are valid real number solutions because when squared, they both return the value 25.

Solving by Square Roots

Solving quadratic equations by taking square roots is a straightforward technique used when the equation can be simplified to \(x^2 = a\). For our example, \(x^2 = 25\), you would take the square root of both sides. This method works best when the quadratic equation is in the form \(x^2 = a\) with no linear component, \(bx\).

Here’s how to proceed:

• Start from the simplified equation \(x^2 = 25\).
• To undo the squaring of \(x\), take the square root on both sides. This step reveals the potential solutions as \(x = \pm \sqrt{25}\).
• The symbol \(\pm\) indicates that there are generally two solutions: one positive and one negative.

After calculating, we find that \(\sqrt{25} = 5\). Thus, the solutions are \(x = 5\) and \(x = -5\). This technique efficiently provides the solutions when dealing with simple quadratic equations.

Factoring Quadratic Equations

Factoring is another essential strategy to solve quadratic equations, especially when an equation isn’t as immediately suitable for the square root method. During this method, we convert the quadratic equation into a product of two binomials. For the equation at hand, \(x^{2} - 25 = 0\), we define it as a "difference of squares."

A difference of squares fits the pattern \(a^2 - b^2\) and can be factored as \((a+b)(a-b)\). Applying this to \(x^{2} - 25 = 0\):

• Recognize that the equation is already structured as the difference of squares.
• Rewrite 25 as a perfect square: \(25 = 5^2\).
• Apply the difference of squares formula to factor: \((x+5)(x-5) = 0\).

Once factored, the solutions can be obtained by setting each binomial equal to zero, resulting in \(x + 5 = 0\) or \(x - 5 = 0\). Solving these gives the solutions \(x = -5\) and \(x = 5\). This factorization not only validates our previous method but provides an alternative solution strategy.