Problem 43 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation using any convenient method. \(x^{2}-2 x+1=0\)

Step-by-Step Solution

Answer

The solutions to the quadratic equation \((x-1)^{2}=9\) are \(x = 4\) and \(x = -2\).

Step 1: Rewrite the Equation

Firstly, rewrite the equation from \((x-1)^2=9\) to \(x - 1 = \pm \sqrt{9}\) because the square root of a number can be either positive or negative. So, \(x - 1 = \pm 3\)

Step 2: Solve for \(x\)

We can now isolate \(x\) on one side by adding 1 to each part of the equation. Doing this gives us two equations: \(x - 1 + 1 = 3 + 1\) and \(x - 1 + 1 = -3 + 1\), which simplify to \(x = 4\) and \(x = -2\) respectively.

Step 3: Present the Solution

The solution to the equation \((x-1)^{2}=9\) is \(x = 4\) or \(x = -2\). This is saying that if we substitute 4 or -2 for \(x\) into the original equation, it would hold true.

Key Concepts

Quadratic Formula

When tackling quadratic equations, the quadratic formula is a common method many students learn. This classic formula — \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) — can solve any quadratic equation of the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are coefficients of the equation, and \(\pm\) indicates there will be two solutions, one where you add the square root term and one where you subtract it.

The beauty of this formula is that it's always applicable, no matter the values of \(a\), \(b\), and \(c\), as long as \(a eq 0\). This means you don't have to worry about the equation being factorable or not. Remember, the term under the square root, \(b^2 - 4ac\), is called the discriminant, and it can tell us a lot about the nature of the roots of the equation without even solving it.

Square Root Method

The square root method is an efficient technique for solving quadratic equations that have been designed to form \(x^2 = n\). When a quadratic equation is presented in this manner, simply taking the square root of both sides will give us \(x = \pm\sqrt{n}\), where \(n\) is a constant.

For the equation \( (x-1)^2 = 9 \), applying the square root on both sides leads to \( x - 1 = \pm\sqrt{9} \), which simplifies to \( x - 1 = \pm3 \). It's crucial to include both the positive and negative square roots to account for the two possible values that, when squared, would return to our original constant.

Factoring Quadratics

Factoring quadratics is another powerful approach for solving these types of equations. When a quadratic equation can be factored into the form \( (x - p)(x - q) = 0 \), we can use the Zero Product Property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This leads to \( x = p \) or \( x = q \).

However, not all quadratic equations can be factored easily, especially if they don't have integer roots. In such cases, other methods such as the quadratic formula or completing the square might be more suitable. Factoring is an excellent first method to try, though, since it often requires the least calculation and can provide a quick answer if the quadratic is nicely factorable.

Algebraic Manipulation

Algebraic manipulation encompasses a variety of techniques used to simplify or rearrange equations to make them easier to solve. It's the basis of solving any equation, whether you're isolating variables, combining like terms, or employing operations like addition, subtraction, multiplication, and division on both sides of an equation.

Starting with the given example \( (x-1)^2 = 9 \), using algebraic manipulation, we first apply the square root to both sides, yielding \( x - 1 = \pm3 \). Then, we isolate \(x\) by adding 1 to each side, ultimately finding the two solutions \(x = 4\) and \(x = -2\). This step (isolating \(x\)) is a fundamental part of algebraic manipulation and showcases the importance of understanding and mastering these basic operations to enhance problem-solving skills in algebra.