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Chapter 1: Problem 75

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

Short Answer

Expert verified
The solution to the equation is \(x = 1\) (a double root).

Step by step solution

01

Identify Quadratic Equation

We are given a quadratic equation of the form \(ax^2 + bx + c = 0\). In this case, the coefficients are: \(a = a\), \(b = -(2a + 1)\), and \(c = (a + 1)\).
02

Apply the Quadratic Formula

The formula for finding the roots of a quadratic equation is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the coefficients \(a\), \(b\), and \(c\) from Step 1 into this formula.
03

Calculate the Discriminant

The discriminant \(\Delta\) is given by the expression \(b^2 - 4ac\). Calculate it using \(b = -(2a + 1)\) and \(c = (a + 1)\).\[\Delta = (-(2a+1))^2 - 4(a)(a+1) = (2a+1)^2 - 4a(a+1).\]
04

Simplify the Discriminant

Expand and simplify the expression for the discriminant:\[(2a+1)^2 = 4a^2 + 4a + 1\]and\[4a(a+1) = 4a^2 + 4a.\]Thus, \[\Delta = 4a^2 + 4a + 1 - (4a^2 + 4a) = 1.\]
05

Solve for x

Since \(\Delta = 1\), substitute back into the quadratic formula:\[x = \frac{-(-(2a+1)) \pm \sqrt{1}}{2a} = \frac{2a+1 \pm 1}{2a}.\]This provides two solutions: \[x_1 = \frac{2a+1 + 1}{2a} = \frac{2a+2}{2a} = 1\]and \[x_2 = \frac{2a+1 - 1}{2a} = \frac{2a}{2a} = 1.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of any quadratic equation. If you have an equation in the form \( ax^2 + bx + c = 0 \), the quadratic formula states that the solutions for \( x \) can be found using:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is derived from completing the square technique, allowing us to handle any quadratic equation, regardless of its complexity.
To use the formula effectively:
  • First, ensure your equation is arranged as \( ax^2 + bx + c = 0 \).
  • Identify the coefficients \( a \), \( b \), and \( c \) from your equation.
  • Substitute these values into the quadratic formula to solve for \( x \).
Understanding and using the quadratic formula becomes essential in cases where quadratic equations do not factor neatly, especially with non-integers or complex numbers.
Discriminant
The discriminant is a component of the quadratic formula that gives us insight into the nature of the roots of a quadratic equation. It is represented by the expression \( b^2 - 4ac \).
The value of the discriminant can tell us:
  • If \( \Delta > 0 \), the equation has two distinct real roots.
  • If \( \Delta = 0 \), there is exactly one real root, also known as a repeated or double root.
  • If \( \Delta < 0 \), the roots are complex or imaginary, meaning they cannot be plotted on a standard real number line.
In our exercise, calculating \( \Delta = 1 \) (where \( \Delta = 4a^2 + 4a + 1 - (4a^2 + 4a) \)) tells us that the quadratic equation has two real roots, which happen to be the same in this case as both are equal to 1. The discriminant thus serves as a prelude to knowing how many solutions one can expect before fully solving the quadratic equation.
Roots of a Quadratic Equation
When we talk about 'roots' of a quadratic equation, we are referring to solutions for \( x \) that make the equation true. Graphically, these roots represent the points where the parabola of the quadratic equation intersects the \( x \)-axis.
To find the roots, we:
  • Calculate or simplify the discriminant \( b^2 - 4ac \) to understand the nature of the roots.
  • Substitute the discriminant and coefficients back into the quadratic formula.
  • Perform the arithmetic to solve for \( x \).
In our given problem, after using the quadratic formula, we determined that both possible values of \( x \) are 1. Hence, the quadratic equation \( ax^2 - (2a+1)x + (a+1) = 0 \) has a single visible intersection at \( x = 1 \), confirming it is a double root, meaning the parabola just touches the \( x \)-axis at this point.

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