Problem 75
Question
Solve each of the equations. $$4 x^{2}=5 x$$
Step-by-Step Solution
Verified Answer
The solutions are \(x = 0\) and \(x = \frac{5}{4}\).
1Step 1: Set the Equation to Zero
To solve the quadratic equation \(4x^2 = 5x\), we first set the equation to zero by subtracting \(5x\) from both sides. This gives us the equation \(4x^2 - 5x = 0\).
2Step 2: Factor the Quadratic
Factor the expression \(4x^2 - 5x\). Begin by factoring out the common factor, which is \(x\), resulting in \(x(4x - 5) = 0\).
3Step 3: Solve Each Factor for Zero
According to the zero-product property, if \(ab = 0\) then either \(a = 0\) or \(b = 0\). Apply this property to the factored equation \(x(4x - 5) = 0\). First, set \(x = 0\). Then, set \(4x - 5 = 0\) and solve for \(x\).
4Step 4: Solve the Linear Equation
Solve \(4x - 5 = 0\). Add 5 to both sides to get \(4x = 5\). Next, divide both sides by 4 to isolate \(x\), obtaining \(x = \frac{5}{4}\).
5Step 5: Identify All Solutions
The solutions derived are \(x = 0\) from Step 3, and \(x = \frac{5}{4}\) from Step 4. Thus, the equation \(4x^2 = 5x\) has two solutions: \(x = 0\) and \(x = \frac{5}{4}\).
Key Concepts
Factoring QuadraticsZero-Product PropertySolving Linear Equations
Factoring Quadratics
When you encounter a quadratic equation like the one in the exercise, your first step is often to factor the equation. This is because factoring transforms it into a product of simpler expressions, making it easier to solve.
To factor a quadratic expression, recognize any common factors that can be taken out. For the example, the term "4x^2" and "5x" both have "x" in common. We can express the equation as:
To factor a quadratic expression, recognize any common factors that can be taken out. For the example, the term "4x^2" and "5x" both have "x" in common. We can express the equation as:
- Start with the expression: \(4x^2 - 5x = 0\)
- Factor out the common factor "x": \(x(4x - 5) = 0\)
Zero-Product Property
The zero-product property is a crucial tool used after you have factored a quadratic equation. It states that if the product of two factors equals zero, then at least one of the factors must be zero. This property is especially helpful because it reduces complex equations to simpler linear equations.
Applying this to our exercise:
Applying this to our exercise:
- We have: \(x(4x - 5) = 0\)
- According to the zero-product property, either \(x = 0\) or \(4x - 5 = 0\).
Solving Linear Equations
Once the quadratic equation is broken down into linear parts using factoring and the zero-product property, you solve the resulting linear equations individually.
Take our equation, for example:
Take our equation, for example:
- We derived two equations: \(x = 0\) and \(4x - 5 = 0\).
- The first equation, \(x = 0\), is already solved; x is clearly zero.
- For the second equation, \(4x - 5 = 0\), isolate \(x\) by adding 5 to both sides yielding \(4x = 5\).
- Finally, divide both sides by 4 to solve for \(x\), giving you \(x = \frac{5}{4}\).
Other exercises in this chapter
Problem 75
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