Problem 39 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(4 x^{2}+16 x-9=0\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2}\) and \(x = -\frac{9}{2}\).

Step 1: Identify the Equation

The given equation is a quadratic equation of the form \(ax^2 + bx + c = 0\). Here, \(a = 4\), \(b = 16\), and \(c = -9\).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \(ax^2+bx+c=0\) is given by \(b^2 - 4ac\). Substitute the values: \(b = 16\), \(a = 4\), \(c = -9\). Calculate the discriminant: \(16^2 - 4 \cdot 4 \cdot (-9) = 256 + 144 = 400\).

Step 3: Determine the Number of Solutions

Since the discriminant is positive (\(400 > 0\)), this means that the quadratic equation has two distinct real solutions.

Step 4: Apply the Quadratic Formula

The solutions for the quadratic equation can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the given values into the formula: \(x = \frac{-16 \pm \sqrt{400}}{8}\).

Step 5: Calculate Solutions

First, calculate \(\sqrt{400}\), which is \(20\). Then solve the equation for the two possible values of \(x\): \(x = \frac{-16 + 20}{8}\) or \(x = \frac{-16 - 20}{8}\).

Step 6: Simplify Solutions

For \(x = \frac{-16 + 20}{8}\), simplify to get \(x = \frac{4}{8} = \frac{1}{2}\). For \(x = \frac{-16 - 20}{8}\), simplify to get \(x = \frac{-36}{8} = -\frac{9}{2}\).

Key Concepts

Discriminant

The discriminant of a quadratic equation is a key part of solving for its roots. It's derived from the general form of a quadratic equation, \( ax^2 + bx + c = 0 \). The discriminant, often denoted as \( \Delta \), is calculated using the formula: \( \Delta = b^2 - 4ac \). This simple expression provides crucial information about the nature of the roots of the equation. Let's examine why:

When you calculate the discriminant, you can determine:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (also called a "double root").
• If \( \Delta < 0 \), the equation has no real solutions, meaning the solutions are complex or imaginary.

The discriminant gives insight without needing to solve the equation completely. In the example, \( b = 16 \), \( a = 4 \), and \( c = -9 \), leading to a discriminant of \( 400 \). Since \( 400 > 0 \), two distinct real roots exist for the given equation.

Quadratic Formula

Once we know a quadratic equation possesses real solutions, we can find them precisely using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula arises from the process of completing the square on the general form \( ax^2 + bx + c = 0 \). It might look complex, but it's straightforward when you break it down:

• The formula encompasses all possible solutions \( (x) \).
• The \( \pm \) sign indicates that there will be two potential values for \( x \): one for addition, and one for subtraction.
• Each part of the formula corresponds directly to the components identified in the equation: \( a \), \( b \), and \( c \).

In our exercise, we already determined \( b^2 - 4ac = 400 \) and know \( a = 4 \) and \( b = 16 \). Plugging these values in, the solutions become \( x = \frac{-16 \pm 20}{8} \). This calculation provides us the two distinct values of \( x \) that satisfy the original equation.

Real Solutions

Finding real solutions of a quadratic equation means identifying where the graph of the equation intersects the \( x \)-axis. These intersections are points where \( y \), or \( ax^2 + bx + c \), equals zero.

When the discriminant is positive, two real solutions exist, indicating the graph of the quadratic function crosses the \( x \)-axis at two points. We've concluded from the discriminant that our specific equation indeed has two real solutions. Using the quadratic formula, these solutions were solved as follows:

• For \( x = \frac{-16 + 20}{8} \): Simplifying gives \( x = \frac{1}{2} \).
• For \( x = \frac{-16 - 20}{8} \): Simplifying results in \( x = -\frac{9}{2} \).

These solutions provide the precise \( x \)-values where the quadratic equation achieves zero, allowing you to understand where the curve intersects the \( x \)-axis, thus verifying the existence of real solutions.