What is the solution set of x2 + 5x + 1 = 0?
The Correct Answer and Explanation is :
To solve the quadratic equation ( x^2 + 5x + 1 = 0 ), we can use the quadratic formula. The general form of a quadratic equation is:
[
ax^2 + bx + c = 0
]
where ( a ), ( b ), and ( c ) are constants. In this case, we have:
- ( a = 1 )
- ( b = 5 )
- ( c = 1 )
The quadratic formula is:
[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
]
Step-by-Step Solution:
- Substitute the values of ( a ), ( b ), and ( c ) into the formula:
[
x = \frac{-5 \pm \sqrt{5^2 – 4(1)(1)}}{2(1)}
]
- Simplify the expression under the square root (the discriminant):
[
x = \frac{-5 \pm \sqrt{25 – 4}}{2}
]
[
x = \frac{-5 \pm \sqrt{21}}{2}
]
- Solve for the two possible values of ( x ):
[
x = \frac{-5 + \sqrt{21}}{2} \quad \text{or} \quad x = \frac{-5 – \sqrt{21}}{2}
]
Final Answer:
The solution set of the quadratic equation ( x^2 + 5x + 1 = 0 ) is:
[
x = \frac{-5 + \sqrt{21}}{2} \quad \text{or} \quad x = \frac{-5 – \sqrt{21}}{2}
]
These are the two real and distinct solutions to the equation.
Explanation:
This is a quadratic equation because the highest power of ( x ) is 2, which means it has at most two solutions. In this case, the discriminant ( b^2 – 4ac = 21 ) is positive, which indicates that there are two distinct real solutions.
The solutions come from the quadratic formula, which ensures that every quadratic equation has a solution, either real or complex. In this example, the solutions are real because the discriminant is positive. The expression ( \sqrt{21} ) is irrational, so the solutions are also irrational numbers.
Thus, the solution set is composed of two irrational numbers, representing the values of ( x ) that satisfy the original equation.