Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 27
Solve. $$ y 2-10 y+24=0 $$
Short Answer
Expert verified
The solutions are \( y = 6 \) and \( y = 4 \).
Step by step solution
01
Identify Equation Type
The given equation is a quadratic equation of the form \( ay^2 + by + c = 0 \), where \( a = 1 \), \( b = -10 \), and \( c = 24 \).
02
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 \((-6) + (-4) = -10\).
03
Write the Factored Equation
Use the numbers found in Step 2 to rewrite the equation: \((y - 6)(y - 4) = 0\).
04
Solve for \( y \)
Set each factor equal to zero and solve for \( y \):\[ y - 6 = 0 \Rightarrow y = 6 \]\[ y - 4 = 0 \Rightarrow y = 4 \].
05
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \((y - 6)(y - 4) = 0\).
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 \(y = 4\).
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.
