Problem 67 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square or by using the quadratic formula. $$ y^{2}=2 y+1 $$

Step-by-Step Solution

Answer

Question: Solve the quadratic equation \(y^2 - 2y + 1 = 0\) and find the value of y. Answer: The value of y is 1.

Step 1: Write the equation in the standard form

Write the given equation in the form of 'y^2 - 2y + 1 = 0'. $$ y^{2}- 2y + 1 = 0 $$

Step 2: Completing the square

To complete the square, add and subtract the square of half of the coefficient of the linear term inside the parentheses: $$ \left(y^2 - 2y + 1\right) = (y -1)^2 $$

Step 3: Solve for y

Set the expression equal to 0 and solve for y: $$ (y - 1)^2 = 0 $$ We can deduce that y = 1 is the solution for our equation. Now, let's solve the same equation using the quadratic formula.

Step 1: Write the equation in the standard form

The equation is already in the standard form: $$ y^{2}- 2y + 1 = 0 $$

Step 2: Identify the coefficients

Identify the coefficients a, b, and c for our given equation: a = 1, b= -2, and c = 1

Step 3: Apply the quadratic formula

Apply the quadratic formula, which is: $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Substituting the coefficients: $$ y = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(1)}}{2(1)} $$

Step 4: Solve for y

Simplify the expression to find the value of y: $$ y = \frac{2 \pm \sqrt{4 - 4}}{2}= \frac{2 \pm \sqrt{0}}{2} $$ Since there is only one solution \(y = \frac{2}{2} = 1\), both methods give the same result.

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations by rearranging the terms into a perfect square trinomial. This makes the equation easier to solve. Here’s how to do it:

To complete the square for the equation \(y^2 - 2y + 1 = 0\), you need to work on the quadratic and linear terms, \(y^2 - 2y\). The goal is to make it a perfect square.

• Take half of the coefficient of the linear term (\(-2\)) and square it. Half of \(-2\) is \(-1\), and \((-1)^2 = 1\).
• Add and subtract this squared value inside the equation: \(y^2 - 2y + 1\), which already happens to be a perfect square trinomial: \((y - 1)^2\).

Now, the equation is \((y - 1)^2 = 0\). You just solve \(y - 1 = 0\), leading to \(y = 1\). This method is especially helpful for gaining insight into transformations and roots.

Quadratic Formula

The quadratic formula is a universal tool used to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). This formula is:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Using the quadratic formula for our equation \(y^2 - 2y + 1 = 0\):

• Identify the coefficients: \(a = 1\), \(b = -2\), \(c = 1\).
• Plug these values into the formula: \(y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(1)}}{2(1)}\).

Solving this:

• Calculate inside the square root: \((-2)^2 - 4 \times 1 \times 1 = 4 - 4 = 0\).
• This simplifies to \(y = \frac{2 \pm \sqrt{0}}{2}\), which results in \(y = 1\).

The quadratic formula is very versatile, providing solutions even when the roots are complex.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). Having the equation in this form is crucial for solving it using methods like completing the square or the quadratic formula.

In our exercise, the equation \(y^2 = 2y + 1\) is not in the standard form. To convert it:

• Subtract \(2y\) and \(1\) from both sides to get \(y^2 - 2y + 1 = 0\).

Now, it's in standard form with \(a = 1\), \(b = -2\), and \(c = 1\).

Understanding standard form is key because it sets up the equation for efficient application of solving strategies. By identifying \(a\), \(b\), and \(c\), one can quickly determine the solution method to use.