Problem 38 - Introductory Algebra for College Students

Question

Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$x^{2}-25=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(x^{2}-25=0\) are \(x=5\) and \(x=-5\).

Step 1: Factor the Equation

To factor the equation \(x^{2}-25=0\), look for two numbers that multiply to -25 and add to 0. This is a difference of squares and it factors into \( (x-5)(x+5)=0.\)

Step 2: Solve for x Values

By using the zero product property that states if 'ab = 0', then 'a' or 'b' must be zero: the solutions are \( 'x=5' \) and \( 'x=-5' \).

Step 3: Checking the Solution

Substitute each of these values into the original equation to check that the solutions are correct. For \(x=5\), \(5^{2}-25=25-25=0\). For \(x=-5\), \((-5)^{2}-25=25-25=0\). So, both x values are correct.

Key Concepts

Difference of Squares

The difference of squares is a unique pattern in algebra that simplifies the factoring process. Whenever you see an equation like \(x^2 - a^2 = 0\), it represents the difference of two perfect squares. This is because \(x^2\) and \(a^2\) are both squares, and when they are subtracted, it forms this special type of expression.
To factor a difference of squares, use the formula:

• \(x^2 - a^2 = (x-a)(x+a)\)

You simply find the square root of each term, \(x\) and \(a\), and create two binomials: one with a subtraction sign and the other with an addition sign.
For example, with the equation \(x^2 - 25 = 0\), you recognize \(x^2\) as a square and \(25\) as \(5^2\). The factorization is \((x-5)(x+5)\). This method efficiently handles equations that fit this pattern, allowing you to solve them more easily.

Zero Product Property

The zero product property is a fundamental principle used to solve equations that involve products. It states that if the product of two expressions equals zero, then at least one of the expressions must be zero. Applying this principle is crucial when solving polynomial equations that have been factored.
In the solved example \((x-5)(x+5) = 0\), we apply the zero product property. You set each factor equal to zero:

• \(x-5=0\)
• \(x+5=0\)

Now, solve these simple equations:

• For \(x-5=0\), adding 5 to both sides gives \(x=5\)
• For \(x+5=0\), subtracting 5 from both sides gives \(x=-5\)

This method ensures you find all potential solutions, as each factor can independently contribute to the product being zero.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \(x\) for which the equation equals zero. A quadratic equation commonly takes the form \(ax^2 + bx + c = 0\). Several methods exist to solve such equations, including factoring, using the quadratic formula, and completing the square.
In this context, factoring is an effective technique when the quadratic can be expressed as a product of binomials, especially in the case of the difference of squares.For example, in the given equation \(x^2 - 25 = 0\), recognizing it as a difference of squares allows for quick factorization:

• The equation is rewritten as \((x-5)(x+5) = 0\).

With the factors found, the zero product property comes into play, letting you solve by setting each binomial to zero and solving for \(x\). Thus, the solutions here are \(x=5\) and \(x=-5\), checked by substitution to confirm their validity. This approach is straightforward and allows for a quick assessment of potential solutions, especially useful when the quadratic is easily factorable.