Problem 50
Question
Solve each equation. $$ 3 s^{2}=-4 s $$
Step-by-Step Solution
Verified Answer
The solutions are \(s = 0\) and \(s = -\frac{4}{3}\).
1Step 1: Set Equation to Zero
First, move all terms to one side of the equation to set it to zero: \[ 3s^2 + 4s = 0 \]
2Step 2: Factor the Equation
Factor out the common term \(s\):\[ s(3s + 4) = 0 \]
3Step 3: Apply the Zero Product Property
Using the zero product property, set each factor equal to zero:1. \( s = 0 \)2. \( 3s + 4 = 0 \)
4Step 4: Solve Each Equation
Solve for \(s\) from each factor:1. From \( s = 0 \), we have \( s = 0 \).2. From \( 3s + 4 = 0 \), solve for \(s\) by rearranging and simplifying: \[ 3s = -4 \ s = -\frac{4}{3} \]
5Step 5: State the Solutions
The solutions to the equation \(3s^2 = -4s\) are:\(s = 0\) and \(s = -\frac{4}{3}\).
Key Concepts
Setting Equation to ZeroFactoringZero Product Property
Setting Equation to Zero
In order to solve quadratic equations, it's important to first set the equation to zero. This means that all terms of the equation are on one side, with the equation equating to zero on the other side. This step is critical because it prepares the equation for factoring or other methods of solving.
For example, given the equation \(3s^2 = -4s\), you need to move the \(-4s\) term to the left side of the equation. You do this by adding \(4s\) to both sides:
For example, given the equation \(3s^2 = -4s\), you need to move the \(-4s\) term to the left side of the equation. You do this by adding \(4s\) to both sides:
- Original equation: \(3s^2 = -4s\)
- Move \(-4s\) to the left: \(3s^2 + 4s = 0\)
Factoring
Factoring is a helpful technique used to simplify and solve quadratic equations. It involves expressing the equation as a product of its factors, which are simpler expressions that multiply to create the original equation.
In the example \(3s^2 + 4s = 0\), you can factor out the common factor, which is \(s\), from each term:
In the example \(3s^2 + 4s = 0\), you can factor out the common factor, which is \(s\), from each term:
- Identify common factor: Both terms \(3s^2\) and \(4s\) have the factor \(s\).
- Factor the equation: \(s(3s + 4) = 0\)
Zero Product Property
The Zero Product Property is a key principle used in solving quadratic equations. It states that if a product of two numbers is zero, then at least one of the numbers must be zero. After setting the equation to zero and factoring, this property is used to find solutions.
When you have factored the equation \(3s^2 + 4s = 0\) into \(s(3s + 4) = 0\), the Zero Product Property allows you to set each factor to zero:
When you have factored the equation \(3s^2 + 4s = 0\) into \(s(3s + 4) = 0\), the Zero Product Property allows you to set each factor to zero:
- \(s = 0\)
- \(3s + 4 = 0\)
- For \(s = 0\), the solution is simply \(s = 0\).
- For \(3s + 4 = 0\), solve for \(s\) by subtracting 4 from both sides and dividing by 3: \(s = -\frac{4}{3}\).
Other exercises in this chapter
Problem 50
Factor. $$ 49 d^{2}-16 $$
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Factor. See Example 7 or Example \(12 .\) $$4 b^{2}+12 b-16$$
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Choose the correct method from Section 6.1 through Section 6.5 and factor completely. $$ 64 p^{3}-27 $$
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Factor out the GCF. $$ 18 r-30 r^{2} $$
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