Certainly! Let's solve the equation [tex]\( x(6 + x) = 1 \)[/tex] step by step.
1. Expand the Equation:
First, distribute [tex]\( x \)[/tex] across the terms inside the parentheses.
[tex]\[
x(6 + x) = 1 \implies 6x + x^2 = 1
\][/tex]
2. Rearrange into a Standard Quadratic Form:
To solve the quadratic equation, we need to get all terms on one side of the equation so it equals zero.
[tex]\[
x^2 + 6x - 1 = 0
\][/tex]
3. Identify the Coefficients:
In the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we identify the coefficients:
[tex]\[
a = 1, \quad b = 6, \quad c = -1
\][/tex]
4. Use the Quadratic Formula:
The quadratic formula is given by
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute the identified coefficients [tex]\( a = 1, b = 6, \)[/tex] and [tex]\( c = -1 \)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-1)}}{2(1)}
\][/tex]
5. Simplify under the Square Root:
Calculate the discriminant [tex]\( \Delta = b^2 - 4ac \)[/tex]:
[tex]\[
\Delta = 6^2 - 4(1)(-1) = 36 + 4 = 40
\][/tex]
So the expression under the square root is [tex]\( 40 \)[/tex], therefore:
[tex]\[
x = \frac{-6 \pm \sqrt{40}}{2}
\][/tex]
6. Simplify Further:
We can simplify [tex]\( \sqrt{40} \)[/tex] as
[tex]\[
\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}
\][/tex]
Substitute back to get:
[tex]\[
x = \frac{-6 \pm 2\sqrt{10}}{2}
\][/tex]
7. Separate the Solutions:
Divide each term in the numerator by the denominator (2):
[tex]\[
x = \frac{-6}{2} \pm \frac{2\sqrt{10}}{2} = -3 \pm \sqrt{10}
\][/tex]
8. Write the Final Solutions:
Therefore, the solutions to the equation [tex]\( x(6 + x) = 1 \)[/tex] are:
[tex]\[
x = -3 + \sqrt{10} \quad \text{and} \quad x = -3 - \sqrt{10}
\][/tex]
These are the detailed steps that lead us to the final answer:
[tex]\[
x = -3 + \sqrt{10} \quad \text{and} \quad x = -3 - \sqrt{10}
\][/tex]
