Problem 51 - A Graphical Approach to College Algebra

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$2 x^{2}-4 x=1$$

Step-by-Step Solution

Answer

The solutions are \( x = 1 \pm \frac{\sqrt{6}}{2} \).

Step 1: Set the Equation to Zero

First, modify the equation to equal zero by subtracting 1 from both sides: \[ 2x^2 - 4x - 1 = 0 \]

Step 2: Identify a, b, and c

Next, identify the coefficients in the quadratic equation \( ax^2 + bx + c = 0 \). Here, \( a = 2 \), \( b = -4 \), and \( c = -1 \).

Step 3: Use the Quadratic Formula

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

Step 4: Compute the Discriminant

Calculate the discriminant using the formula \( b^2 - 4ac \).\(-4^2 - 4(2)(-1) = 16 + 8 = 24\). The discriminant is positive, indicating two real solutions.

Step 5: Solve for x

Substitute the discriminant back into the quadratic formula:\[ x = \frac{-(-4) \pm \sqrt{24}}{2(2)} \]Simplify under the square root \( \sqrt{24} = 2\sqrt{6} \), thus:\[ x = \frac{4 \pm 2\sqrt{6}}{4} \]Which simplifies to:\[ x = 1 \pm \frac{\sqrt{6}}{2} \]

Step 6: Graphical Verification

Plot the function \( f(x) = 2x^2 - 4x - 1 \) on a graph. The points where the graph intersects the x-axis correspond to the roots found: \( x = 1 + \frac{\sqrt{6}}{2} \) and \( x = 1 - \frac{\sqrt{6}}{2} \). These confirm the two solutions found.

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). It's especially useful when factoring is complex or impossible. The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it works:

• The "\( b^2 - 4ac \)" part under the square root is called the discriminant. It tells us about the nature of the roots.
• The "\( -b \pm \)" denotes that there can be two potential solutions: one by adding the square root part and one by subtracting it.
• Divide the result by \( 2a \), which is twice the coefficient of \( x^2 \).

These steps give us the complete set of solutions for the quadratic equation, making it a reliable method for finding the values of \( x \) that satisfy the equation. In our example, we applied these steps to get the solutions \( x = 1 \pm \frac{\sqrt{6}}{2} \). This means we have two real and distinct solutions for the equation.

discriminant

The discriminant is a key element of the quadratic formula. It is given by \( D = b^2 - 4ac \) and plays a crucial role in determining the nature of the roots of a quadratic equation. Here's how it helps:

• If \( D > 0 \), there are two distinct real roots. This means the graph of the quadratic equation will intersect the x-axis at two points.
• If \( D = 0 \), there is exactly one real root, indicating that the graph touches the x-axis at one point (a "perfect square").
• If \( D < 0 \), there are no real roots. The graph does not intersect the x-axis at all, and instead, the solutions are complex numbers.

In our example problem, the discriminant was calculated to be 24, which is positive. This indicates there are two real solutions for the quadratic equation \( 2x^2 - 4x - 1 = 0 \). Thus, the equation has two points where it crosses the x-axis.

graphical solution

Finding a graphical solution means plotting the equation to visually identify its roots. For a quadratic equation, this involves drawing its parabola and observing where it crosses the x-axis. Each intersection point corresponds to a real solution for the equation.For the equation \( 2x^2 - 4x - 1 \), plot the function to see its shape:

• The highest-degree term \( 2x^2 \) indicates that the graph opens upwards since the coefficient is positive.
• The vertex form or axis of symmetry gives us a rough idea of where the graph is centered. You can find this by \( x = -\frac{b}{2a} \).

When we plot the equation, the graphical solution shows points of intersection at \( x = 1 + \frac{\sqrt{6}}{2} \) and \( x = 1 - \frac{\sqrt{6}}{2} \). These intersections confirm the real solutions obtained mathematically. Graphical verification is a great way to ensure your calculations are correct and to visualize the problem solution.

real solutions

Real solutions to a quadratic equation are the x-values where the parabola crosses the x-axis. For an equation like \( 2x^2 - 4x - 1 = 0 \), if the discriminant is positive, it assures us of real and distinct solutions.Here's what this tells us:

• Real solutions are straightforward to find using calculators or graphing tools, as these methods verify the exact points of intersection.
• In our problem, solutions \( x = 1 + \frac{\sqrt{6}}{2} \) and \( x = 1 - \frac{\sqrt{6}}{2} \) are those real solutions. You can also solve visually by observing the graph.
• Real solutions also imply tangible, valid answers in most practical real-world scenarios. This can relate to problems of areas, velocities, and more. When applying these in real situations, ensure measurements and computations reflect reality.

Understanding the concept of real solutions enriches the understanding of the mathematical model and fits it into practical applications, allowing us to solve problems accurately with confidence.