Certainly! I'll walk you through the steps needed to solve the quadratic equation [tex]\(x^2 + x = 4\)[/tex].
1. Rewrite the equation in standard quadratic form:
The standard form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex].
Starting with
[tex]\[
x^2 + x = 4,
\][/tex]
we need to rearrange this equation to make it equal to zero:
[tex]\[
x^2 + x - 4 = 0.
\][/tex]
2. Identify the coefficients:
In the equation [tex]\(x^2 + x - 4 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex] (coefficient of [tex]\(x^2\)[/tex]),
- [tex]\(b = 1\)[/tex] (coefficient of [tex]\(x\)[/tex]),
- [tex]\(c = -4\)[/tex] (constant term).
3. Apply the quadratic formula:
The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\][/tex]
Substituting the identified coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula:
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}.
\][/tex]
4. Simplify inside the square root:
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 16}}{2}.
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{17}}{2}.
\][/tex]
5. Express the final solutions:
The term inside the square root is [tex]\(17\)[/tex], which does not simplify further. Thus, we have two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-1 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{17}}{2}.
\][/tex]
Using these steps, we find that the solutions to the equation [tex]\(x^2 + x = 4\)[/tex] are:
[tex]\[
x = \frac{-1 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{17}}{2}.
\][/tex]
Or more succinctly:
[tex]\[
x = -\frac{1}{2} + \frac{\sqrt{17}}{2} \quad \text{and} \quad x = -\frac{\sqrt{17}}{2} - \frac{1}{2}.
\][/tex]
