Problem 71 - Algebra and Trigonometry

Question

Solve the equation for \(x\). \(a x^{2}-(2 a+1) x+(a+1)=0 \quad(a \neq 0)\)

Step-by-Step Solution

Answer

The solutions are \(x = 1 + \frac{1}{a}\) and \(x = 1\).

Step 1: Identify the Equation

The given equation is a quadratic equation in the form \(ax^2 + bx + c = 0\). Here, we have \(a = a\) (coefficient of \(x^2\)), \(b = -(2a+1)\) (coefficient of \(x\)), and \(c = a+1\).

Step 2: Apply the Quadratic Formula

The quadratic formula states that for a quadratic equation \(ax^2 + bx + c = 0\), the solutions for \(x\) are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plugging in \(a = a\), \(b = -(2a+1)\), and \(c = a+1\), we get: \[ x = \frac{-( -(2a+1)) \pm \sqrt{(-(2a+1))^2 - 4 imes a imes (a+1)}}{2a} \]

Step 3: Simplify Within the Formula

Calculate the discriminant \(b^2 - 4ac\):- \(b^2 = (-(2a+1))^2 = (2a+1)^2 = 4a^2 + 4a + 1\)- \(4ac = 4a(a+1) = 4a^2 + 4a\)Thus, the discriminant is \(b^2 - 4ac = 4a^2 + 4a + 1 - (4a^2 + 4a) = 1\).

Step 4: Compute the Solutions

Substitute the discriminant back into the quadratic formula:\[ x = \frac{2a + 1 \pm \sqrt{1}}{2a} \]Since \(\sqrt{1} = 1\), the solutions for \(x\) become:\[ x = \frac{2a + 1 \pm 1}{2a} \]

Step 5: Find Distinct Solutions

Calculate the two possible values for \(x\):1. \(x = \frac{2a + 1 + 1}{2a} = \frac{2a + 2}{2a} = \frac{a + 1}{a} = 1 + \frac{1}{a}\)2. \(x = \frac{2a + 1 - 1}{2a} = \frac{2a}{2a} = 1\)The solutions are \(x = 1 + \frac{1}{a}\) and \(x = 1\).

Key Concepts

Quadratic Formula

In solving quadratic equations, the quadratic formula is a powerful tool that provides a systematic way to find the solutions. The quadratic formula is expressed as follows:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula is especially useful when factoring isn't doable or obvious. The parameters \(a\), \(b\), and \(c\) represent coefficients from the standard quadratic equation of the form \(ax^2 + bx + c = 0\).
Plugging these coefficients into the formula allows you to directly compute the values of \(x\), the roots of the quadratic equation, by utilizing elementary arithmetic operations and square roots.
It handles any quadratic equation, ensuring the discovery of up to two solutions. By following this standard, errors associated with manual attempts at factoring are minimized. This makes the quadratic formula an ideal choice for solving equations where coefficients are not straightforward.

Discriminant

An essential aspect of the quadratic formula is the discriminant. It is nestled inside the square root of the quadratic formula:\[D = b^2 - 4ac\]The discriminant provides key insights into the nature and number of solutions a quadratic equation might have.

• If \(D > 0\), there are two distinct real solutions.
• If \(D = 0\), there is exactly one real solution, also known as a repeated or double root.
• If \(D < 0\), the solutions are complex numbers and not real.

Understanding the discriminant helps hypothesize about the solution without fully solving it initially, letting you know what kind of roots to expect.
In the problem given, we computed \(b^2 - 4ac = 1\), leading to \(D > 0\), indicating two real solutions are possible, which we confirm in the subsequent steps.

Solutions of a Quadratic Equation

Upon computing the discriminant and applying the quadratic formula, the next step is to find the actual solutions of the equation. These solutions represent the values of \(x\) that satisfy the given equation, serving as points where the related quadratic graph intersects the x-axis.
For our exercise, substituting the known values into the quadratic formula gave us both:

• \(x = 1 + \frac{1}{a}\)
• \(x = 1\)

These represent the two distinct roots or solutions of the quadratic equation we've been working on.
The process entails plugging in the values into the formula, simplifying appropriately, and deriving the applicable roots. This step underscores the utility of both the quadratic formula and discriminant, showcasing their combined power in yielding solutions efficiently.
This approach guarantees precision and can be systematically applied to any quadratic equation with certainty, leading to accurate results every single time.