91Ó°ÊÓ

The quadratic formula is a powerful tool for solving quadratic equations. Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The standard form ensures every term is aligned properly, with the highest degree term (\( x^2 \)) first.
The quadratic formula itself is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, the term \( b^2 - 4ac \) is crucial. It is called the discriminant, and it helps determine the nature of the roots.
When the discriminant is positive, the equation has two distinct real solutions. If it's zero, there is exactly one real solution. On the other hand, a negative discriminant means no real solutions, but you can find complex ones. In our context, we substitute the specific values of \( a \), \( b \), and \( c \) to find the exact solutions for our quadratic equation.

To solve quadratic equations, we use the quadratic formula. The first step is to plug in the values of \( a \), \( b \), and \( c \) from the equation into the formula. For the given equation, \( x^2 + 10x + 2 = 0 \), we identified that \( a = 1 \), \( b = 10 \), and \( c = 2 \).
Substituting these into the formula gives us:

• \( x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \)

This simplifies the problem as we compute inside the square root first. Calculating the discriminant \( 10^2 - 4 \cdot 1 \cdot 2 \) gives 92. So, our equation reduces to:

• \( x = \frac{-10 \pm \sqrt{92}}{2} \)

Solving equations after this involves simplifying further to find the precise values for \( x \). The methodical substitution into the quadratic formula systematically brings us closer to the solutions.

Simplifying radicals is a task often encountered when using the quadratic formula. When we reach
a stage where we are required to take the square root of a number, simplifying the radical simplifies the work. In our problem, to simplify \( \sqrt{92} \), we identify its prime factors.
The number 92 factors into 2 and 23, specifically \( 92 = 4 \times 23 \). Since 4 is a perfect square, we can simplify \( \sqrt{92} \) as \( \sqrt{4 \times 23} = 2\sqrt{23} \).
This reduced radical reflects a more straightforward form of the expression and grows our understanding of number properties.
Simplifying radicals yields an easier expression to work with when continuing to solve for \( x \). In practice, this means replacing \( \sqrt{92} \) with \( 2\sqrt{23} \), and subsequently substituting back into our equation, we can then solve for \( x \) easily:

• \( x = \frac{-10 \pm 2\sqrt{23}}{2} \)

Ultimately, it results in our final solutions: \( x = -5 + \sqrt{23} \) and \( x = -5 - \sqrt{23} \). Being able to simplify radicals efficiently allows mathematics to be more manageable and less cumbersome.