Problem 113 - Elementary Algebra

Question

Solve by using the Quadratic Formula. \(v(v+5)-10=0\)

Step-by-Step Solution

Answer

The solutions are \( v = \frac{-5 + \sqrt{65}}{2} \) and \( v = \frac{-5 - \sqrt{65}}{2} \).

Step 1: Expand the Equation

Start by expanding the equation. Given: \(v(v+5) - 10 = 0\). Expand the first term: \(v^2 + 5v - 10 = 0\)

Step 2: Write in Standard Form

Ensure that the equation is in the standard form of a quadratic equation \(ax^2 + bx + c = 0\). Your equation is already in standard form: \(v^2 + 5v - 10 = 0\)

Step 3: Identify Coefficients

Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. Here, \(a = 1\), \(b = 5\), and \(c = -10\)

Step 4: Apply the Quadratic Formula

The Quadratic Formula is: \[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substitute the values of \(a\), \(b\), and \(c\) into the formula: \[ v = \frac{-5 \pm \sqrt{5^2 - 4(1)(-10)}}{2(1)} \]

Step 5: Solve Under the Square Root

Calculate the discriminant (the part under the square root): \[5^2 - 4(1)(-10) = 25 + 40 = 65\] Therefore, the formula becomes: \[ v = \frac{-5 \pm \sqrt{65}}{2} \]

Step 6: Simplify the Results

Split the equation into the two possible solutions for \(v\): \[ v_1 = \frac{-5 + \sqrt{65}}{2} \] \[ v_2 = \frac{-5 - \sqrt{65}}{2} \]. These are the two solutions for the equation.

Key Concepts

Expanding Equations

To solve quadratic equations, one fundamental skill is expanding equations. In the given problem, we start with the equation: \(v(v+5) -10 = 0\). Expanding involves distributing the terms inside the parentheses. Multiply \(v\) by each term inside the parentheses: \(v \times v + v \times 5\), which results in \(v^2 + 5v\). This gives us: \(v^2 + 5v - 10 = 0\). Expanding helps simplify and prepare the equation for further steps.

Standard Form of a Quadratic Equation

A quadratic equation should be in standard form for many solving techniques. The standard form is: \(ax^2 + bx + c = 0\). Here, 'a', 'b', and 'c' are coefficients, and 'x' is the variable. After expanding, our equation \(v^2 + 5v - 10 = 0\) is already in this form. In this equation:

• \(a = 1\) (the coefficient of \(v^2\))
• \(b = 5\) (the coefficient of \(v\))
• \(c = -10\) (the constant term)

Ensuring an equation is in standard form is essential before proceeding with methods like the Quadratic Formula.

Discriminant in Quadratic Equations

The discriminant helps us understand the nature of the solutions for a quadratic equation. It is found within the Quadratic Formula: \(b^2 - 4ac\). For our problem, the discriminant is calculated as follows:

• \(b = 5\)
• \(a = 1\)
• \(c = -10\)

Plug these values into the discriminant formula: \(5^2 - 4(1)(-10) = 25 + 40 = 65\). The discriminant here is \(65\). Since it is positive, we have two distinct real solutions.

Solving Quadratics

Solving a quadratic equation typically involves finding values of the variable that make the equation true. We use the Quadratic Formula: \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting our coefficients \(a = 1\), \(b = 5\), and \(c = -10\):

• First calculate the discriminant: \(65\).
• Plug the discriminant and coefficients into the formula: \(v = \frac{-5 \pm \sqrt{65}}{2}\).

This gives us two potential solutions:

• \(v_1 = \frac{-5 + \sqrt{65}}{2}\)
• \(v_2 = \frac{-5 - \sqrt{65}}{2}\)

These values satisfy the original equation, illustrating the powerful method of the Quadratic Formula in solving quadratic equations.