Problem 107 Find the real solutions, if any,... [FREE SOLUTION] (2024)

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The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \). The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is used to find the roots (solutions) of the equation.

In our given problem, we first converted it into standard form: \( 3m^2 + 6m + 1 = 0 \). Then, we identified the coefficients: \( a = 3 \), \( b = 6 \), and \( c = 1 \). Using the quadratic formula, we can directly plug in these values:
\( m = \frac{-6 \pm \sqrt{36 - 12}}{6} \).

This part of solving quadratics is particularly useful because it works for every quadratic equation, regardless of whether the roots are real or complex.

The discriminant is a critical part of the quadratic formula and helps us determine the nature of the roots of the quadratic equation. The discriminant is found using the formula: \( \ \text{Discriminant} = b^2 - 4ac \).

If the discriminant:

• is greater than zero: the quadratic equation has two distinct real roots.
• is equal to zero: the quadratic equation has exactly one real root.
• is less than zero: the quadratic equation has two complex roots, and no real solutions.

In our problem, we computed the discriminant as follows: \( \ b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot 1 = 36 - 12 = 24 \).

Since our discriminant is positive (\( 24 \ \text{is greater than 0} \)), we know there are two distinct real solutions.

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating these symbols. It allows us to express and solve equations using variables, like those seen in quadratic equations.

For example, we started with the equation: \( 3m^2 + 6m + 1 = 0 \). Using algebra, we rearranged it into the standard quadratic form and identified the coefficients. We then applied algebraic principles to simplify our results after using the quadratic formula.

This involved breaking down the square root term, simplifying fractions, and arriving at two potential solutions:
\( m = -1 + \frac{\sqrt{6}}{3} \) \(\text{and} \-1 - \frac{\sqrt{6}}{3} \).

Knowing basic algebraic operations helps in solving more complex problems by building up from simpler steps.