Question
Solve the quadratic by factoring.
x^2−38=−3x−10
x^2−38=−3x−10
Answers
Answered by
GPT-4o mini
To solve the quadratic equation \(x^2 - 38 = -3x - 10\), we first rearrange the equation to set it to zero.
Start by moving all terms to one side:
\[
x^2 + 3x - 38 + 10 = 0
\]
This simplifies to:
\[
x^2 + 3x - 28 = 0
\]
Next, we need to factor the quadratic \(x^2 + 3x - 28\). We need to find two numbers that multiply to \(-28\) (the constant term) and add up to \(3\) (the coefficient of \(x\)).
After testing possible pairs, we find that \(7\) and \(-4\) work:
\[
7 \cdot (-4) = -28 \quad \text{and} \quad 7 + (-4) = 3
\]
Thus, we can factor the quadratic as follows:
\[
(x + 7)(x - 4) = 0
\]
Next, we set each factor to zero to solve for \(x\):
1. \(x + 7 = 0 \Rightarrow x = -7\)
2. \(x - 4 = 0 \Rightarrow x = 4\)
Therefore, the solutions to the equation \(x^2 - 38 = -3x - 10\) are:
\[
\boxed{-7} \quad \text{and} \quad \boxed{4}
\]
Start by moving all terms to one side:
\[
x^2 + 3x - 38 + 10 = 0
\]
This simplifies to:
\[
x^2 + 3x - 28 = 0
\]
Next, we need to factor the quadratic \(x^2 + 3x - 28\). We need to find two numbers that multiply to \(-28\) (the constant term) and add up to \(3\) (the coefficient of \(x\)).
After testing possible pairs, we find that \(7\) and \(-4\) work:
\[
7 \cdot (-4) = -28 \quad \text{and} \quad 7 + (-4) = 3
\]
Thus, we can factor the quadratic as follows:
\[
(x + 7)(x - 4) = 0
\]
Next, we set each factor to zero to solve for \(x\):
1. \(x + 7 = 0 \Rightarrow x = -7\)
2. \(x - 4 = 0 \Rightarrow x = 4\)
Therefore, the solutions to the equation \(x^2 - 38 = -3x - 10\) are:
\[
\boxed{-7} \quad \text{and} \quad \boxed{4}
\]