Problem 71 - Algebra A Combined Function

Question

Solve each linear or quadratic equation \(2 x^{2}-x-1=0\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Equation Type

The given equation is \(2x^2 - x - 1 = 0\). This is a quadratic equation because it has the \(x^2\) term. The standard form for a quadratic equation is \(ax^2 + bx + c = 0\), where \(a, b, c\) are constants. Here, \(a = 2\), \(b = -1\), and \(c = -1\).

Step 2: Use the Quadratic Formula

For a quadratic equation in the form \(ax^2 + bx + c = 0\), the solutions can be found using the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). In our case, \(a = 2\), \(b = -1\), and \(c = -1\). Substitute these values into the formula.

Step 3: Calculate the Discriminant

The discriminant in the quadratic formula is \(b^2 - 4ac\). Substitute \(b = -1\), \(a = 2\), and \(c = -1\), so \(b^2 - 4ac = (-1)^2 - 4(2)(-1) = 1 + 8 = 9\). The discriminant is 9, which is positive, indicating there are two real and distinct solutions.

Step 4: Solve for x using the Quadratic Formula

Substitute the values and discriminant into the quadratic formula: \(x = \frac{{-(-1) \pm \sqrt{9}}}{{2(2)}}\). This simplifies to \(x = \frac{{1 \pm 3}}{4}\).

Step 5: Find the Two Solutions

Calculate the two solutions: \(x = \frac{{1 + 3}}{4} = 1\) and \(x = \frac{{1 - 3}}{4} = -\frac{1}{2}\).

Key Concepts

Understanding the Quadratic Formula

Quadratic equations, like the one given in the problem, are expressions that include an \(x^2\) term. These are best solved using the quadratic formula, which helps find values of \(x\) that make the equation equal zero. The quadratic formula is:

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

This might look complicated, but it's just a tool to simplify finding the solutions of any equation structured as \(ax^2 + bx + c = 0\). In this formula, \(a\), \(b\), and \(c\) are coefficients from the equation. By substituting these values into the formula, we ensure that nothing is missed and every solution can be correctly calculated.
When you approach a quadratic equation, always verify its form. Ensure it’s in the standard form which is usually \(ax^2 + bx + c = 0\). Identifying such a form is your first step before using the quadratic formula effectively.

What is the Discriminant?

The discriminant is part of the quadratic formula located under the square root: \(b^2 - 4ac\). This number plays a crucial role in determining the type and nature of the solutions of the quadratic equation.

• If the discriminant is positive, as in our example where it equals 9, this tells you that the equation has two distinct real solutions, meaning there are two different solutions for \(x\).
• If it equals zero, it means there is exactly one real solution, or a repeated solution. The parabola represented by the quadratic equation will just touch the x-axis at one point.
• If it is negative, this indicates there are no real solutions, only complex ones, because you can't take the square root of a negative number without moving into imaginary numbers.

So, always start by calculating the discriminant. It will guide you on what to expect, and it can even save time when you know how many solutions you will need to compute.

Interpreting Real and Distinct Solutions

For the quadratic equation we solved, since the discriminant was positive, there are two real and distinct solutions. These solutions signify points where the parabola (graph of the quadratic equation) cuts the x-axis. Each solution corresponds to one of these intersections.
Let's imagine the graph visually:

• When you see two points where the parabola intersects the x-axis, you have real and distinct solutions. This happens when the parabola opens upwards or downwards without just touching or hovering above the axis.
• If there was only one point of intersection, it implies the parabola is just grazing the x-axis at its vertex. This would indicate a single, repeated solution.
• If it does not touch the x-axis at all, it reaffirms no real solutions exist; the parabola sits completely above or below the x-axis.

In our problem, the two solutions \(x = 1\) and \(x = -\frac{1}{2}\) tell us exactly where the parabola crosses the x-axis. Understanding the nature and number of solutions can also help in visualizing and graphing quadratic equations accurately.