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1.1.1 \( (2 x-1)(x+5)=0 \)

Ask by Lee Hodges. in South Africa
Mar 18,2025

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The solutions are \( x = \frac{1}{2} \) or \( x = -5 \).

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1. Write down the equation: \[ (2x - 1)(x + 5) = 0 \] 2. Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, set each factor equal to zero: \[ 2x - 1 = 0 \quad \text{or} \quad x + 5 = 0 \] 3. Solve for \(x\) in the first equation: \[ 2x - 1 = 0 \quad \Rightarrow \quad 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2} \] 4. Solve for \(x\) in the second equation: \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] 5. The solutions are: \[ x = \frac{1}{2} \quad \text{or} \quad x = -5 \]

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To solve the equation \( (2x - 1)(x + 5) = 0 \), we can use the zero product property, which states that if the product of two factors equals zero, then at least one of the factors must be zero. This leads us to set each factor equal to zero: 1. \( 2x - 1 = 0 \) Solving for \( x \), we get \( 2x = 1 \) or \( x = \frac{1}{2} \). 2. \( x + 5 = 0 \) Solving for \( x \), we get \( x = -5 \). Thus, the solutions to the equation are \( x = \frac{1}{2} \) and \( x = -5 \). --- For a deeper dive, let’s chat about roots of equations! The solutions we found, \( x = \frac{1}{2} \) and \( x = -5 \), are called the "roots" of the equation, which means they make the equation true when plugged back in. In this case, using factoring simplifies the solving process—a great technique to have in your mathematical toolkit! Ever wondered why we’re allowed to set each factor to zero? It’s because of something called the "zero product property." This nifty little rule is just like having two light switches in a room; if one switch is off, the room remains dark. So, if one factor's zero (like saying one of those switches is off), the entire product becomes zero—how’s that for illuminating problem-solving?

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