Problem 68 - Algebra 2

Question

What are the solutions of the equation \(6 x^{2}+9 x-15=0 ?\) $$ \begin{array}{ll}{\text { F. } 1,-15} & {\text { G. } 1,-\frac{5}{2}} \\\ {\text { H. }-1,-5} & {\text { J. } 3, \frac{5}{2}}\end{array} $$

Step-by-Step Solution

Answer

The solutions of the equation are x = 3/2 and x = -5/3.

Step 1: Write down the given quadratic equation

Begin by writing down the given quadratic equation, which is 6x^2 + 9x - 15 = 0.

Step 2: Factor the quadratic equation

Factor the quadratic equation if possible. For the equation 6x^2 + 9x - 15 = 0, it can be factored into (2x - 3)(3x + 5) = 0.

Step 3: Solve for x

Since the equation is factored, set each factor equal to zero: (2x - 3) = 0 and (3x + 5) = 0. Then, solve each equation for x.

Step 4: Calculate the solutions

Solving the first equation (2x - 3) = 0 gives x = 3/2. Solving the second equation (3x + 5) = 0 gives x = -5/3.

Key Concepts

Factoring Quadratic Equations

Factoring is a key step in solving quadratic equations, which are algebraic equations in the form of \(ax^2+bx+c=0\) where \(a\), \(b\), and \(c\) are constants and \(a\) is not equal to zero.

In the simplified process of factoring, we look for two numbers that multiply to give the product of \(a\) and \(c\) and add to give \(b\). For the equation \(6x^2+9x-15=0\), we need to find factors of \(-90\) (since \(6\times{-15}=-90\)) that add up to \(9\). These numbers are \(15\) and \(-6\). We can then rewrite the equation as \(6x^2+15x-6x-15=0\) and group the terms to factor by grouping. This step is crucial as it transforms the equation into a format where the Quadratic Formula is not necessary and the equation can be solved through simpler means.

Solving Quadratic Equations

Once a quadratic equation is factored, solving it becomes much more straightforward. After having factored our example to \((2x - 3)(3x + 5)=0\), the next step is to apply the Zero Product Property. This property tells us that if a product of two factors equals zero, at least one of the factors must be zero.

That leads to setting each factor equal to zero: \(2x - 3=0\) and \(3x + 5=0\), which can be solved to find the solutions for \(x\). Solving each equation for \(x\) gives us the solutions \(x = 3/2\) and \(x = -5/3\), respectively. It's important to check each solution by substituting them back into the original equation to ensure they truly zero out the equation.

Algebraic Equations

An algebraic equation is a mathematical statement indicating that two expressions are equal. The equations can include variables, constants, and operational symbols. In the case of quadratic equations, they involve a variable raised to the second power. These equations can model various real-world scenarios, such as projectile motion and area problems.

Algebraic equations are solved by first simplifying each side of the equation as much as possible, then using various methods such as factoring, the quadratic formula, or completing the square to find the values of the variables that make the equation true. Understanding not just how to operate with algebraic equations but also why each step is taken contributes greatly to problem-solving skills.

Quadratic Formula

When Factoring is Not Feasible

There are instances where factoring a quadratic equation directly is impractical or impossible. This is where the Quadratic Formula comes in. This formula is derived from completing the square and can solve any quadratic equation. The standard formula is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), where \(a\), \(b\), and \(c\) are the same constants from the general form of the quadratic equation \(ax^2+bx+c=0\).

While our original equation could be factored easily, knowing the Quadratic Formula is a valuable tool for equations that resist simple factorization. As an essential part of the mathematician's toolkit, it guarantees a method to find the roots of any quadratic equation, provided one is comfortable with using the square root operation and managing the presence of possible complex numbers when the discriminant (\(b^2 - 4ac\)) is negative.