Problem 27 - Elementary Algebra

Question

Solve. $$ y 2-10 y+24=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = 6 \) and \( y = 4 \).

Step 1: Identify Equation Type

The given equation is a quadratic equation of the form \( ay^2 + by + c = 0 \), where \( a = 1 \), \( b = -10 \), and \( c = 24 \).

Step 2: Factor the Quadratic

Look for two numbers that multiply to \( c = 24 \) and add up to \( b = -10 \). These numbers are \(-6\) and \(-4\) because \((-6) \times (-4) = 24\) and \((-6) + (-4) = -10\).

Step 3: Write the Factored Equation

Use the numbers found in Step 2 to rewrite the equation: \((y - 6)(y - 4) = 0\).

Step 4: Solve for \( y \)

Set each factor equal to zero and solve for \( y \):\[ y - 6 = 0 \Rightarrow y = 6 \]\[ y - 4 = 0 \Rightarrow y = 4 \].

Step 5: Verify the Solutions

Substitute \( y = 6 \) and \( y = 4 \) back into the original equation to verify they satisfy it:For \( y = 6 \):\( 6^2 - 10(6) + 24 = 36 - 60 + 24 = 0 \).For \( y = 4 \):\( 4^2 - 10(4) + 24 = 16 - 40 + 24 = 0 \).Both solutions satisfy the original equation.

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations is a method used to express the quadratic equation in a product form. This product form helps to easily find the solutions of the equation. For the given exercise, the quadratic equation is in the form \( y^2 - 10y + 24 = 0 \). The goal of factoring is to rewrite this equation as \((y - m)(y - n) = 0\). Here are the steps to factor:

• Identify Constants: Recognize the constants \(a = 1\), \(b = -10\), and \(c = 24\).


• Find Two Numbers: Look for two numbers whose product is \(c = 24\) and whose sum is \(b = -10\). The numbers are \(-6\) and \(-4\) because \((-6) \,\times\, (-4) = 24\) and \((-6) + (-4) = -10\).


• Write in Factored Form: Use these numbers to express the equation as \((y - 6)(y - 4) = 0\).

Factoring turns the quadratic equation into a form that can be easily solved by setting each factor equal to zero and solving for \(y\). This makes it a crucial step in handling quadratic equations.

Solving Quadratic Equations

Once the quadratic equation is factored into two binomials, solving it becomes straightforward. For the equation \((y - 6)(y - 4) = 0\), we need to find the values of \(y\) that make the equation true. To do this, we use the zero-product property:

• Set Each Factor to Zero: This property states that if the product of two things equals zero, at least one of them must be zero. Thus, set each factor to zero:
  • \(y - 6 = 0\)
  • \(y - 4 = 0\)


• Solve for \(y\): Solving these simple linear equations gives the solutions:
  • For \(y - 6 = 0\), adding 6 to both sides yields \(y = 6\).
  • For \(y - 4 = 0\), adding 4 to both sides gives \(y = 4\).

The solutions to the equation are \(y = 6\) and \(y = 4\). These values of \(y\) satisfy the factored form of the equation, proving they are correct solutions to the original quadratic equation.

Verifying Solutions

Verifying solutions ensures that the values found for \(y\) using the zero-product property satisfy the original quadratic equation. This step is vital to confirm the accuracy of the solutions. Here's how to verify:

• Substitute Solutions: Substitute each solution back into the original equation to see if it results in a true statement:
  • For \(y = 6\): Calculate \(6^2 - 10(6) + 24 = 36 - 60 + 24\), which simplifies to \(0\).
  • For \(y = 4\): Calculate \(4^2 - 10(4) + 24 = 16 - 40 + 24\), which also simplifies to \(0\).


• Confirm Zero Result: Both substitutions result in zero, confirming that \(y = 6\) and \(y = 4\) are valid solutions.

Verification checks the consistency and correctness of your solutions. It is a simple yet powerful step that eliminates errors and reinforces understanding of the solving process for quadratic equations.