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Topic 2/3
15 Flashcards in this deck.
A quadratic equation takes the standard form: $$ ax^2 + bx + c = 0 $$ where \( a \), \( b \), and \( c \) are coefficients with \( a \neq 0 \). The solutions to this equation are the values of \( x \) that satisfy it. Quadratic equations graph as parabolas, which either open upwards or downwards depending on the sign of \( a \).
Factorization involves expressing the quadratic equation as a product of two binomials. This method is effective when the equation can be easily factored.
Step-by-Step Process:
Example:
Solve \( 2x^2 + 5x + 3 = 0 \) by factorization.
1. Multiply \( a \) and \( c \): \( 2 \times 3 = 6 \).
2. Find numbers that multiply to 6 and add to 5: 2 and 3.
3. Rewrite the equation: \( 2x^2 + 2x + 3x + 3 = 0 \).
4. Factor by grouping: \( 2x(x + 1) + 3(x + 1) = 0 \).
5. Factor out \( (x + 1) \): \( (2x + 3)(x + 1) = 0 \).
6. Solve: \( 2x + 3 = 0 \) leads to \( x = -\frac{3}{2} \); \( x + 1 = 0 \) leads to \( x = -1 \).
The quadratic formula provides a universal method for solving any quadratic equation: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ This formula derives from completing the square and offers solutions even when factorization is complex or impossible.
Example:
Solve \( 1x^2 - 4x + 4 = 0 \) using the quadratic formula.
Here, \( a = 1 \), \( b = -4 \), and \( c = 4 \). Plugging into the formula: $$ x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)} = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2 $$ So, the equation has one real solution: \( x = 2 \).
Graphic display calculators (GDCs) are powerful tools for solving quadratic equations by visualizing their graphs. By plotting the quadratic equation, students can identify the points where the parabola intersects the x-axis, representing the solutions.
Steps to Solve:
Example:
For the equation \( x^2 - 5x + 6 = 0 \), input the function \( y = x^2 - 5x + 6 \) into the GDC. The graph will display a parabola intersecting the x-axis at \( x = 2 \) and \( x = 3 \), indicating the solutions.
Understanding the graph of a quadratic equation enhances comprehension of its solutions. The vertex form of a quadratic equation is: $$ y = a(x - h)^2 + k $$ where \( (h, k) \) is the vertex of the parabola. Analyzing the vertex and the direction of the parabola provides insights into the nature of the solutions.
Discriminant Analysis:
The discriminant \( D = b^2 - 4ac \) determines the nature of the roots:
The quadratic formula is derived by completing the square on the general quadratic equation \( ax^2 + bx + c = 0 \).
Step-by-Step Derivation:
This derivation underscores the reliability and universality of the quadratic formula in solving any quadratic equation.
When the discriminant \( D = b^2 - 4ac \) is negative, the solutions to the quadratic equation are complex numbers. These solutions can be expressed as: $$ x = \frac{-b}{2a} \pm \frac{\sqrt{4ac - b^2}}{2a}i $$ where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).
Example:
Solve \( x^2 + 4x + 8 = 0 \).
Here, \( a = 1 \), \( b = 4 \), and \( c = 8 \). The discriminant \( D = 16 - 32 = -16 \). Thus, $$ x = \frac{-4 \pm \sqrt{-16}}{2} = \frac{-4 \pm 4i}{2} = -2 \pm 2i $$ So, the solutions are \( x = -2 + 2i \) and \( x = -2 - 2i \).
Quadratic equations model various real-life scenarios, such as projectile motion, optimization problems, and geometric configurations.
Projectile Motion:
The height \( h \) of an object thrown upwards with initial velocity \( v_0 \) is given by: $$ h = v_0 t - \frac{1}{2}gt^2 $$ where \( g \) is the acceleration due to gravity. Solving for \( t \) involves quadratic equations.
Optimization:
Maximizing area or minimizing cost often leads to quadratic equations. For instance, finding the dimensions that maximize the area of a rectangular enclosure given a fixed perimeter.
Quadratic equations intersect with various disciplines:
Challenging quadratic problems may require combining methods or applying them in non-standard contexts.
Example:
Find the values of \( x \) that satisfy both \( x^2 + y^2 = 25 \) and \( y = x + 1 \).
Substitute \( y = x + 1 \) into the first equation: $$ x^2 + (x + 1)^2 = 25 \\ x^2 + x^2 + 2x + 1 = 25 \\ 2x^2 + 2x - 24 = 0 \\ x^2 + x - 12 = 0 $$ Factorizing: $$ (x + 4)(x - 3) = 0 $$ Thus, \( x = -4 \) or \( x = 3 \). Corresponding \( y \) values are \( y = -3 \) and \( y = 4 \), respectively.
Method | Definition | Applications | Pros | Cons |
---|---|---|---|---|
Factorization | Expressing the quadratic as a product of binomials. | Simple equations with easy factors. | Quick and straightforward when applicable. | Not feasible for complex or non-factorable equations. |
Quadratic Formula | A universal formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. | All quadratic equations, regardless of their nature. | Always provides solutions, including complex ones. | Can be cumbersome for simple equations. |
Graphic Display Calculator | Using technology to graph and find intersections. | Visualizing solutions and understanding graph behavior. | Provides visual insights and handles complex equations. | Requires access to a calculator and familiarity with its functions. |
Memorize the Quadratic Formula: Keep \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) handy for quick reference.
Check Your Solutions: Always substitute your answers back into the original equation to verify correctness.
Organize Your Work: Clearly label each step when solving to avoid errors and make it easier to review.
Use the Discriminant: Quickly determine the nature of the roots by calculating \( D = b^2 - 4ac \) before solving.
Did you know that quadratic equations have been studied for over 4000 years? Ancient Babylonian mathematicians were among the first to solve quadratic equations, primarily for practical purposes like land division and architecture. Additionally, quadratic equations play a crucial role in modern technology, including computer graphics and engineering design, where they help model curves and optimize structures.
Mistake 1: Forgetting to set the quadratic equation to zero before solving.
Incorrect: Solving \( 2x^2 + 5x + 3 = 4 \) directly.
Correct: First, subtract 4 to get \( 2x^2 + 5x - 1 = 0 \).
Mistake 2: Incorrectly calculating the discriminant.
Incorrect: Using \( D = b^2 + 4ac \) instead of \( D = b^2 - 4ac \).
Correct: Always use \( D = b^2 - 4ac \) to determine the nature of the roots.
Mistake 3: Misapplying the quadratic formula by forgetting the negative sign.
Incorrect: \( x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a} \).
Correct: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).