Question
Solve the equation \( x^{2}+3 x-28=0 \) by factoring. \[ x=\square \]
Ask by Hilton Ingram. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solutions are \( x = -7 \) or \( x = 4 \).
Solution
We start with the quadratic equation
\[
x^2 + 3x - 28 = 0.
\]
To factor, we look for two numbers that multiply to \(-28\) and add to \(3\). These numbers are \(7\) and \(-4\) because
\[
7 \times (-4) = -28 \quad \text{and} \quad 7 + (-4) = 3.
\]
Thus, we can factor the equation as
\[
(x+7)(x-4)=0.
\]
Now, we set each factor equal to zero:
1. \(x+7=0\) implies
\[
x=-7.
\]
2. \(x-4=0\) implies
\[
x=4.
\]
Therefore, the solutions are
\[
x=-7 \quad \text{or} \quad x=4.
\]
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Simplify this solution Extra Insights
To solve the equation \( x^{2}+3 x-28=0 \) by factoring, we need to find two numbers that multiply to \(-28\) (the constant term) and add up to \(3\) (the coefficient of the linear term). The numbers \(7\) and \(-4\) fit the bill since \(7 \times -4 = -28\) and \(7 + (-4) = 3\). Now we can write the equation as: \[ (x + 7)(x - 4) = 0 \] Setting each factor equal to zero gives us: \[ x + 7 = 0 \quad \text{or} \quad x - 4 = 0 \] From \(x + 7 = 0\), we get \(x = -7\), and from \(x - 4 = 0\), we get \(x = 4\). Thus, the solutions are: \[ x = -7 \quad \text{or} \quad x = 4 \]
