Problem 37 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ t^{2}-25=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Equation

The equation given is a quadratic equation because it is in the form of \( ax^2 + bx + c = 0 \). In this case, \( a=1 \), \( b=0 \), and \( c=-25 \).

Step 2: Rearrange the Equation

Rewrite the equation to make it easier to solve. The given equation is already simple: \( t^2 = 25 \).

Step 3: Take the Square Root of Both Sides

To solve for \( t \), we take the square root of both sides of the equation: \( t = \pm \sqrt{25} \).

Step 4: Simplify the Square Root

Calculate the square root of 25. \( \sqrt{25} \) is 5, so \( t = \pm 5 \).

Step 5: List the Solutions

The solutions to the equation are \( t = 5 \) and \( t = -5 \).

Key Concepts

Solving Quadratic Equations

Quadratic equations are essential in algebra and appear in the form \( ax^2 + bx + c = 0 \). These equations are characterized by the presence of \( x^2 \), and the first step in solving them is understanding what each variable represents. Here, \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant term. The equation \( t^{2} - 25 = 0 \) is already simplified, making it easier to solve.

• First, we recognize that this is a quadratic equation with \( a = 1 \), \( b = 0 \), and \( c = -25 \).
• Next, we attempt to simplify the process by trying various methods like factoring or square roots.
• Often, the method chosen depends on the form of the equation and which technique will provide the simplest solution path.

Understanding these concepts is crucial for approaching more complex equations efficiently.

Square Root Method

The square root method is particularly useful when the quadratic equation is in the form \( x^2 = n \), where n is a constant. In such scenarios, solving for the unknown can be straightforward by directly applying the square root.Taking the square root cancels out the exponent on \( x \). Here is how you apply it:

• Isolate the squared term if necessary. For \( t^2 - 25 = 0 \), we rewrite it as \( t^2 = 25 \).
• Apply the square root to both sides, which gives you \( t = \pm \sqrt{25} \).
• Thus, simplifying \( \sqrt{25} \) gives \( t = \pm 5 \), leading to the solutions: \( t = 5 \) and \( t = -5 \).

This method is simple and effective, especially for equations missing the \( x \) term or those that can be made into perfect squares.

Factoring Quadratic Equations

Factoring is another powerful tool in solving quadratic equations, especially when they can be rewritten or arranged as a pattern of binomials multiplying each other.When considering a simple expression \( t^2 - 25 = 0 \), it is recognized as a difference of squares because it fits into the pattern \( a^2 - b^2 = (a - b)(a + b) \).

• The expression \( t^2 - 25 \) can be factored into \( (t - 5)(t + 5) = 0 \).
• By setting each factor to zero, \( t - 5 = 0 \) and \( t + 5 = 0 \), we can solve for \( t \).
• This gives the solutions \( t = 5 \) and \( t = -5 \).

Factoring is efficient for equations that exhibit this pattern, making it an excellent technique for quickly finding solutions without complex calculations.