Problem 137 - Algebra and Trigonometry

Question

Will help you prepare for the material covered in the next section. $$ \text { Solve: } 2 x^{2}+x=15 $$

Step-by-Step Solution

Answer

The roots of the quadratic equation are \(x = 2.5\) and \(x = -3\).

Step 1: Rearrange the Equation

Rearrange the quadratic equation in the standard form \(ax^2 + bx + c = 0\). In our case, we subtract 15 from both sides to get \(2x^2 + x - 15 =0\). The coefficients are \(a=2, b=1, c=-15\).

Step 2: Use Quadratic Formula

Now, apply the quadratic formula to solve for X. Using \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute the expressed coefficients into the formula.

Step 3: Calculation

Substituting our coefficients in, we get: \[x = \frac{-1 \pm \sqrt{(1)^2 - 4*2*-15}}{2*2}\]. Upon simplifying, we get \(x = \frac{-1 \pm \sqrt{121}}{4}\). This further simplifies to \[x = \frac{-1 \pm 11}{4}\].

Step 4: Find Roots

Solving for \(x\) we get: \(x = \frac{-1 + 11}{4}\) or \(x = \frac{-1 - 11}{4}\). Hence the roots of the equation are \(x=2.5\) or \(x = -3\).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in solving quadratic equations. It's particularly useful when factoring is not straightforward. The formula itself is given by:

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

This formula works for any quadratic equation of the form \( ax^2 + bx + c = 0 \). Notice that it includes several key components:

• \(-b\): This term accounts for the opposite of the linear coefficient \(b\).
• \(\pm \): Represents the two potential solutions stemming from the equation.
• \(\sqrt{b^2 - 4ac}\): Known as the discriminant, it determines the type and number of solutions.
• \(2a\): Indicates the division, ensuring the equation remains balanced.

Always remember to check the discriminant (\(b^2 - 4ac\)) as it will give you insight into the nature of the roots. If it's positive, you'll find two distinct real roots. If zero, there's one real root, and if negative, the roots are complex.

Standard Form of a Quadratic

A quadratic equation should be in the standard form before applying the quadratic formula. The standard form is:

• \( ax^2 + bx + c = 0 \)

This format ensures clarity in identifying the coefficients \(a\), \(b\), and \(c\) necessary for solving. Let's break it down:

• \(a\): The coefficient of \(x^2\), indicating how the parabola opens up (when positive) or down (when negative).
• \(b\): The coefficient of \(x\), important in determining the axis of symmetry.
• \(c\): The constant term, representing the y-intercept of the quadratic graph.

Rearranging an equation into standard form is often the first step in solving quadratics. For example, our equation \(2x^2 + x = 15\) was rearranged by subtracting 15, giving \(2x^2 + x - 15 = 0\). This sets the stage for applying the quadratic formula.

Finding Roots Using Quadratic Formula

Once the quadratic equation is in its standard form, finding its roots becomes straightforward with the quadratic formula. Let’s look back at our example:

• Start by identifying the coefficients, \(a = 2\), \(b = 1\), \(c = -15\).
• Substitute these into the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Applying it:

• \( x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot -15}}{2 \cdot 2} \)
• This simplifies to \( x = \frac{-1 \pm \sqrt{121}}{4} \), where 121 is the perfect square 11 squared.

Solving further, we find:

• \( x = \frac{-1 + 11}{4} = 2.5 \)
• \( x = \frac{-1 - 11}{4} = -3 \)

Thus, the roots of the equation \(2x^2 + x = 15\) are \(x = 2.5\) and \(x = -3\). Understanding this process allows you to solve any quadratic, given the coefficients of its standard form.