Problem 19 - Elementary Algebra

Question

Solve by factoring and then solve using the quadratic formula. Check answers. $$y 2-2 y+1=0$$

Step-by-Step Solution

Answer

The solution is \( y = 1 \).

Step 1: Rewrite the Equation

First, let's rewrite the quadratic equation: \( y^2 - 2y + 1 = 0 \). This is a standard quadratic equation where \( a = 1 \), \( b = -2 \), and \( c = 1 \).

Step 2: Factor the Equation

Observe that the quadratic can be rewritten as \( (y - 1)^2 = 0 \). This is because \( (y - 1)(y - 1) = y^2 - 2y + 1 \).

Step 3: Solve by Factoring

From the factorized form \( (y - 1)^2 = 0 \), set \( y - 1 = 0 \) to find the solution. This gives us \( y = 1 \).

Step 4: Solve Using the Quadratic Formula

Recall the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = -2 \), and \( c = 1 \) into the formula: \[ y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \].

Step 5: Calculate the Discriminant

Compute the discriminant (\( b^2 - 4ac \)): \((-2)^2 - 4 \cdot 1 \cdot 1 = 4 - 4 = 0 \).

Step 6: Solve the Formula

Since the discriminant is 0, we have a single solution: \[ y = \frac{2 \pm 0}{2} = \frac{2}{2} = 1 \].

Step 7: Verify the Solutions

Both methods have yielded the same solution, \( y = 1 \). Substitute \( y = 1 \) back into the original equation: \( 1^2 - 2 \cdot 1 + 1 = 1 - 2 + 1 = 0 \), confirming it satisfies the equation.

Key Concepts

Factoring

Factoring is a method used to solve quadratic equations by expressing the quadratic expression as a product of two linear factors. This technique simplifies the equation, allowing you to find its roots easily. For the quadratic equation given, we observed that it was possible to rewrite it from its original form, \( y^2 - 2y + 1 \), to a perfect square: \((y - 1)^2\).

Recognizing perfect squares is crucial in factoring because it reduces the work required to solve the equation. In simple terms, factoring boils down to finding two numbers that multiply to the constant term (in this case, 1) and add to the middle coefficient (-2). Practicing factoring regularly helps build an intuition for judging whether a quadratic can be simplified this way, which saves time and effort in solving equations.

Quadratic Formula

The quadratic formula is a universal method for finding the solutions to any quadratic equation. It is given by:

• \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula is derived from the process of completing the square and provides a reliable way to solve quadratics when factoring is cumbersome or not possible. Applying this formula involves substituting the coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).

For our exercise, the quadratic formula helped us verify the solution found via factoring. By using the formula with \( a = 1 \), \( b = -2 \), and \( c = 1 \), we demonstrated that \( y = 1 \) is indeed the solution. This method is particularly useful when roots are not integers or when decimals are involved.

Discriminant

The discriminant is a component of the quadratic formula located under the square root, \( b^2 - 4ac \). It provides important information about the nature of the roots of a quadratic equation:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If negative, there are no real roots, but two complex roots instead.

In our exercise, the discriminant calculated to be zero, indicating that both methods—factoring and using the quadratic formula—would yield exactly one solution, which was \( y = 1 \). Understanding the discriminant gives insight into what kind of solutions to expect before actually solving the equation.

Solving Equations

Solving quadratic equations involves finding values of \( y \) that satisfy the equation. Typically, this begins with identifying which method is most suitable—factoring, the quadratic formula, or completing the square.

Each method may be preferred depending on the specific equation:

• Factoring works best when the quadratic is easily reducible to factors.
• The quadratic formula is great for any quadratic equation, ensuring a solution exists even when factoring isn't apparent.

In this exercise, both methods affirmed that the solution was \( y = 1 \). By checking the solution through substitution into the original equation, we verified accuracy. Solving quadratics goes beyond finding the root; it is also about understanding the properties of the equation and confirming that solutions meet the equation's conditions.