Problem 19 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 4 x^{2}+4 x-1=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-1+\sqrt{2}}{2}\) and \(x = \frac{-1-\sqrt{2}}{2}\).

Step 1 - Identify coefficients

To apply the quadratic formula, the equation must be in the standard form of \[ax^2 + bx + c = 0\]. Identify the coefficients: \(a = 4\), \(b = 4\), and \(c = -1\).

Step 2 - Write down the quadratic formula

The quadratic formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

Step 3 - Substitute the coefficients into the formula

Substitute \(a = 4\), \(b = 4\), and \(c = -1\) into the quadratic formula: \[x = \frac{-(4) \pm \sqrt{(4)^2 - 4(4)(-1)}}{2(4)}\].

Step 4 - Simplify inside the square root

Calculate the value inside the square root: \[(4)^2 - 4(4)(-1) = 16 + 16 = 32\]. The equation now is: \[x = \frac{-4 \pm \sqrt{32}}{8}\].

Step 5 - Simplify the square root

Simplify \(\sqrt{32}\): \[\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}\]. The equation now is: \[x = \frac{-4 \pm 4\sqrt{2}}{8}\].

Step 6 - Simplify the expression

Factor out a 4 from the numerator: \[x = \frac{4(-1 \pm \sqrt{2})}{8} = \frac{-1 \pm \sqrt{2}}{2}\].

Key Concepts

quadratic equation

To solve a quadratic equation, it often helps to identify its standard form. A quadratic equation is any equation that can be written as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable we need to solve for.
In this form:

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

Quadratic equations are used in many fields, such as physics, engineering, and business. They can model various scenarios like projectiles in motion and areas of geometric shapes.

real number solutions

When we work with quadratic equations, the solutions can be real or complex numbers. Real numbers solutions occur when we find real values for \(x\) that satisfy the equation.
The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is key here. The term under the square root, called the discriminant, is \(b^2 - 4ac\). This discriminant determines the nature of the solutions:

• If \(b^2 - 4ac > 0\), we get two distinct real number solutions.
• If \(b^2 - 4ac = 0\), we get exactly one real number solution, also known as a repeated root.
• If \(b^2 - 4ac < 0\), there are no real number solutions; instead, we get complex solutions.

In our exercise, because all solutions are real numbers, we focus on problems where the discriminant (\(b^2 - 4ac\)) is non-negative.

solving quadratic equations

To solve quadratic equations using the quadratic formula, follow these detailed steps:
1. Ensure the equation is in standard form \(ax^2 + bx + c = 0\). Identify \(a\), \(b\), and \(c\).
2. Write down the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
3. Substitute the values of \(a\), \(b\), and \(c\) into the formula.
In our example where \(4x^2 + 4x - 1 = 0\):

• \(a = 4\)
• \(b = 4\)
• \(c = -1\)

4. Calculate the discriminant \(b^2 - 4ac\):
\[4^2 - 4 \times 4 \times -1 = 16 + 16 = 32\]
5. Substitute and simplify: \[x = \frac{-4 \pm \sqrt{32}}{8}\]
Further simplifying: \[\sqrt{32} = 4\sqrt{2}\], thus \[ x = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2}\]
This gives the two real number solutions for \(x\).