Problem 17
Question
Solve and check each of the equations. \(3 x(x-10)+80=5\)
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 5\).
1Step 1: Expand the Expression
First, expand the expression on the left side of the equation: \[ 3x(x - 10) + 80 = 5 \] Use the distributive property: \[ 3x^2 - 30x + 80 = 5 \]
2Step 2: Simplify and Rearrange the Equation
Move all terms to one side to form a quadratic equation: \[ 3x^2 - 30x + 80 - 5 = 0 \] Simplify: \[ 3x^2 - 30x + 75 = 0 \]
3Step 3: Factor the Quadratic
Factor the quadratic equation if possible. We can factor by looking for two numbers that multiply to 225 (\(3*75\)) and add to -30. Factoring gives:\[ 3(x - 5)(x - 5) = 0 \] This simplifies to: \[ 3(x - 5)^2 = 0 \]
4Step 4: Solve for x
Since the equation is \(3(x - 5)^2 = 0\): Divide by 3:\[ (x - 5)^2 = 0 \] Take the square root of both sides: \[ x - 5 = 0 \] So, \(x = 5\).
5Step 5: Verify the Solution
Substitute \(x = 5\) back into the original equation:\[ 3(5)(5 - 10) + 80 = 5 \] Calculate:\[ 3(5)(-5) + 80 = 5 \] \[ -75 + 80 = 5 \] \[ 5 = 5 \] The solution satisfies the original equation.
Key Concepts
Distributive PropertyFactoring QuadraticsSolving Equations Step by StepVerifying Solutions
Distributive Property
The distributive property is a fundamental concept in algebra, making it easier to multiply a single term by terms inside a set of parentheses. It's like handing out each item in a basket individually to those in the group.
For instance, in the equation given, the expression is initially \( 3x(x-10) + 80 = 5 \). To clear the parentheses, we apply the distributive property, multiplying the term outside the parentheses, 3x, by each term inside the parentheses:
For instance, in the equation given, the expression is initially \( 3x(x-10) + 80 = 5 \). To clear the parentheses, we apply the distributive property, multiplying the term outside the parentheses, 3x, by each term inside the parentheses:
- First multiply: \( 3x \times x = 3x^2 \)
- Then multiply: \( 3x \times (-10) = -30x \)
Factoring Quadratics
Factoring quadratics involves breaking down a quadratic equation into simpler expressions that can be multiplied to give the original equation. It's a bit like unwrapping and then neatly folding a large sheet into smaller squares.
In the simplified equation \( 3x^2 - 30x + 75 = 0 \), we aim to factor it. Here we look for two numbers that multiply to 225 (the product of the coefficient of \( x^2 \) and the constant term, i.e., \(3\times75\)) and add up to -30 (the coefficient of \( x \)).
In the simplified equation \( 3x^2 - 30x + 75 = 0 \), we aim to factor it. Here we look for two numbers that multiply to 225 (the product of the coefficient of \( x^2 \) and the constant term, i.e., \(3\times75\)) and add up to -30 (the coefficient of \( x \)).
- In this case, both numbers are -15 and -15 because \(-15 \times -15 = 225\) and \(-15 + -15 = -30\).
Solving Equations Step by Step
Solving equations step by step ensures that each action taken leads you closer to finding the correct solution. Start by transforming all terms into a readable format and simplify where possible.
For our equation, once factored as \( 3(x - 5)^2 = 0 \), we simplify it by dividing by the coefficient of the factored expression, 3, leading to \( (x - 5)^2 = 0 \).
To solve for \( x \), take the square root of both sides, resulting in \( x - 5 = 0 \). This implies that \( x = 5 \). Working step by step allows us to ensure each part of the equation is handled methodically, avoiding mistakes and yielding a reliable solution.
For our equation, once factored as \( 3(x - 5)^2 = 0 \), we simplify it by dividing by the coefficient of the factored expression, 3, leading to \( (x - 5)^2 = 0 \).
To solve for \( x \), take the square root of both sides, resulting in \( x - 5 = 0 \). This implies that \( x = 5 \). Working step by step allows us to ensure each part of the equation is handled methodically, avoiding mistakes and yielding a reliable solution.
Verifying Solutions
Verifying your solution is like double-checking your work to ensure accuracy. This step builds confidence in your final answer. Once we have \( x = 5 \), substitute it back into the original equation, \( 3x(x-10) + 80 = 5 \), to ensure the result holds true.
By substituting \( x \) with 5:
By substituting \( x \) with 5:
- Calculate: \( 3(5)(5-10) + 80 \)
- This simplifies to: \( 3(5)(-5) + 80 \)
- Which becomes: \(-75 + 80 = 5 \)
Other exercises in this chapter
Problem 17
In \(15-26,\) solve each inequality and write the solution set if the variable is an element of the set of integers. $$ |b+6| \leq 5 $$
View solution Problem 17
Write the solution set of each inequality if x is an element of the set of integers. \(2 x^{2}-2 x-24>0\)
View solution Problem 17
Perform the indicated operations and write the result in simplest form. \((y-1)\left(y^{2}-2 y+1\right)\)
View solution Problem 17
Use the definition of subtraction to write each subtraction as a sum. \(-2-5=-7\)
View solution