Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution. $$ 2 x^{2}-8 x-5=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 2 + \frac{\sqrt{26}}{2} \) and \( x = 2 - \frac{\sqrt{26}}{2} \).

Step 1: Identify Coefficients

First, identify the coefficients for the quadratic equation in the form \( ax^2 + bx + c = 0 \). Here, \( a = 2 \), \( b = -8 \), and \( c = -5 \).

Step 2: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 2 \), \( b = -8 \), and \( c = -5 \) into the formula.

Step 3: Calculate the Discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac \). Substitute the values to get \( \Delta = (-8)^2 - 4 \times 2 \times (-5) = 64 + 40 = 104 \).

Step 4: Evaluate the Roots

Since the discriminant \( \Delta = 104 \) is positive, we will have two real roots. Calculate them using the quadratic formula: \[ x = \frac{-(-8) \pm \sqrt{104}}{2 \times 2} \] \[ x = \frac{8 \pm \sqrt{104}}{4} \] Simplify \( \sqrt{104} = 2\sqrt{26} \) to find the exact roots: \[ x = \frac{8 \pm 2\sqrt{26}}{4} \] Which simplifies further to \[ x = 2 \pm \frac{\sqrt{26}}{2} \].

Key Concepts

Understanding the Quadratic Formula

When you're trying to solve a quadratic equation like \(2x^2 - 8x - 5 = 0\), one of the most reliable methods is using the quadratic formula. The equation for this formula is:

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

This tool is crucial, especially when factoring isn't direct or possible. The quadratic formula can find the roots (or solutions) of any quadratic equation. Just remember to plug in the correct coefficients from your equation into the formula, and you will be able to find the solutions.
It is a powerful formula because it works for all kinds of quadratic equations, ensuring you can find the roots whether they are real or complex. Let's explore what these coefficients are because they play an essential role in using the quadratic formula effectively.

Coefficients in Quadratic Equations

In quadratic equations such as \(ax^2 + bx + c = 0\), the letters \(a\), \(b\), and \(c\) are known as coefficients. These numbers tell you how much each term, \(x^2\), \(x\), and the constant (with no \(x\)), contribute to the equation.

• \(a\) is the coefficient of \(x^2\).
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term.

These coefficients are crucial in determining the shape of the parabola described by the quadratic equation and in finding solutions using the quadratic formula.
For the equation \(2x^2 - 8x - 5 = 0\), we identified:

• \(a = 2\)
• \(b = -8\)
• \(c = -5\)

By substituting these into the quadratic formula, you can solve for \(x\) to find the points where the parabola crosses the x-axis.

The Role of the Discriminant

The discriminant is a component of the quadratic formula that can tell you a lot about the roots of a quadratic equation without solving it completely. It is represented by the part under the square root: \(b^2 - 4ac\).
The value of the discriminant helps you in predicting the nature of the roots:

• If the discriminant is positive, you have two distinct real roots.
• If it is zero, there is exactly one real root, also known as a double root.
• If negative, no real roots exist—only complex ones.

In our equation, the discriminant turned out to be \(104\) (since \(64 + 40 = 104\)), which is positive. This means there are two real solutions to the equation \(2x^2 - 8x - 5 = 0\). Understanding the discriminant at this step can guide you on what type of answers to expect.

Finding the Real Roots

Once you've established that the discriminant is positive, indicating two real roots, the next step is calculating these roots. Using the quadratic formula:

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

You substitute the values of \(a\), \(b\), and \(c\) into the equation. For \(2x^2 - 8x - 5 = 0\), this means:

• \(x = \frac{-(-8) \pm \sqrt{104}}{4}\)
• \(x = \frac{8 \pm \sqrt{104}}{4}\)
• Simplifying, we get \(x = 2 \pm \frac{\sqrt{26}}{2}\).

By performing these calculations, you will find the exact points where the parabola crosses the x-axis. These are your real roots, highlighting the solution to the equation.