- Identify the coefficients \(a = 1, b = 6, c = 2\) from the equation \(x^2 + 6x + 2 = 0\).
- Calculate the discriminant \(\Delta = b^2 - 4ac = 36 - 8 = 28\).
- Apply the quadratic formula \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\) to find \(x = \frac{-6 \pm 2\sqrt{7}}{2}\).
- Simplify to obtain the two exact solutions \(\boxed{-3 + \sqrt{7}}\) and \(\boxed{-3 - \sqrt{7}}\).
Explanation
- Analysis of the Quadratic Equation
The objective is to solve the quadratic equation \(x^2 + 6x + 2 = 0\). This is a second-degree polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In this specific equation, we identify the coefficients as \(a = 1\), \(b = 6\), and \(c = 2\). The standard method to solve such an equation is using the quadratic formula, which is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula will provide the roots, which are the values of \(x\) that satisfy the equation.
- Calculating the Discriminant
First, we calculate the discriminant, denoted by \(\Delta\), which is the part of the quadratic formula under the square root, \(\Delta = b^2 - 4ac\). Substituting the identified coefficients into this expression, we get:
\[\Delta = (6)^2 - 4(1)(2)\]
\[\Delta = 36 - 8 = 28\]
The discriminant is positive (\(\Delta > 0\)), which confirms that there are two distinct real roots for this equation.
- Solving and Simplifying the Roots
Now, we substitute the values of \(a\), \(b\), and the discriminant into the quadratic formula:
\[x = \frac{-6 \pm \sqrt{28}}{2(1)}\]
To simplify the expression, we note that \(\sqrt{28}\) can be factored as \(\sqrt{4 \times 7} = 2\sqrt{7}\). Thus, the formula becomes:
\[x = \frac{-6 \pm 2\sqrt{7}}{2}\]
Dividing both terms in the numerator by the denominator, we find:
\[x = -3 \pm \sqrt{7}\]
This results in two exact solutions: \(x_1 = -3 + \sqrt{7}\) and \(x_2 = -3 - \sqrt{7}\). These are the simplified roots of the given quadratic equation.
Examples
Quadratic equations are fundamental in various real-life scenarios, such as modeling the trajectory of a projectile. If a ball is thrown into the air, its height \(h\) at any time \(t\) can be modeled by a quadratic equation, \(h(t) = -gt^2 + vt + h_0\). Solving for when the height is zero (\(h(t)=0\)) allows us to determine precisely when the object hits the ground, helping scientists and engineers predict the landing position and time of flight in sports physics or aviation.