Methods of Solving Quadratic Equations
Solving quadratic equations is a fundamental skill in high school math. There are several methods, but the most common are factorization and the quadratic formula. These methods allow us to find the roots (or solutions) of a quadratic equation, which represent the x-values where the parabola defined by the equation intersects the x-axis. Choosing the right method depends on the specific equation. Factorization is often easier when possible, while the quadratic formula always works.
Formulae
Here are the key formulas:
- General Form of a Quadratic Equation: $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a \neq 0$.
- Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
- Factorization: Expressing the quadratic expression as a product of two linear factors (e.g., $(x + p)(x + q) = 0$). To solve, set each factor to zero and solve for x.
Examples
Example-1: Factorization
Solve $x^2 + 5x + 6 = 0$ using factorization.
We need to find two numbers that add up to 5 (the coefficient of x) and multiply to 6 (the constant term). These numbers are 2 and 3.
Therefore, we can factor the equation as $(x + 2)(x + 3) = 0$.
Setting each factor to zero gives us:
- $x + 2 = 0 \Rightarrow x = -2$
- $x + 3 = 0 \Rightarrow x = -3$
The solutions are $x = -2$ and $x = -3$.
Example-2: Quadratic Formula
Solve $2x^2 – 7x + 3 = 0$ using the quadratic formula.
Here, $a = 2$, $b = -7$, and $c = 3$.
Applying the formula:
$x = \frac{-(-7) \pm \sqrt{(-7)^2 – 4(2)(3)}}{2(2)}$
$x = \frac{7 \pm \sqrt{49 – 24}}{4}$
$x = \frac{7 \pm \sqrt{25}}{4}$
$x = \frac{7 \pm 5}{4}$
This gives us two solutions:
- $x = \frac{7 + 5}{4} = \frac{12}{4} = 3$
- $x = \frac{7 – 5}{4} = \frac{2}{4} = \frac{1}{2}$
The solutions are $x = 3$ and $x = \frac{1}{2}$.
Common mistakes by students
- Incorrectly identifying coefficients: Mixing up the values of a, b, and c in the quadratic formula is a frequent error. Always double-check that the coefficients are correctly matched to the equation.
- Sign errors: Carelessly handling negative signs, especially in the quadratic formula and during simplification, leads to wrong answers. Always be meticulous with signs.
- Forgetting the $\pm$ in the quadratic formula: The quadratic formula provides two solutions (unless the discriminant is zero). Forgetting to calculate both solutions is a common mistake.
- Misunderstanding factorization: Not knowing how to factor a quadratic expression can lead to incorrect answers or the need to use the quadratic formula unnecessarily.
- Arithmetic errors: Simple arithmetic mistakes during the calculation of the discriminant ($b^2 – 4ac$) or the final solution.
Real Life Application
Quadratic equations have many real-world applications, including:
- Projectile Motion: Calculating the trajectory of objects thrown in the air (e.g., a ball, a rocket) involves quadratic equations.
- Engineering: Designing bridges, buildings, and other structures often uses quadratic equations to determine the optimal shapes and sizes.
- Optimization Problems: Finding the maximum or minimum values in a given situation, such as maximizing profit or minimizing cost, can often be modeled using quadratic equations.
- Financial Modeling: Calculating compound interest, investment growth, and other financial scenarios.
Fun Fact
The quadratic formula has been known for thousands of years! Evidence suggests that mathematicians in ancient Babylonia (around 2000 BCE) were already solving quadratic equations.
Recommended YouTube Videos for Deeper Understanding
Q.1 What are the solutions to the quadratic equation $x^2 – 5x + 6 = 0$?
Check Solution
Ans: A
Factoring the quadratic equation, we get $(x – 2)(x – 3) = 0$. Therefore, $x = 2$ or $x = 3$.
Q.2 Using the quadratic formula, find the solutions of the equation $2x^2 + 7x + 3 = 0$.
Check Solution
Ans: A
Applying the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$, where $a = 2$, $b = 7$, and $c = 3$. Thus, $x = \frac{-7 \pm \sqrt{7^2 – 4(2)(3)}}{2(2)} = \frac{-7 \pm \sqrt{49 – 24}}{4} = \frac{-7 \pm \sqrt{25}}{4} = \frac{-7 \pm 5}{4}$. This yields $x = -1/2$ or $x = -3$.
Q.3 Which of the following is a factor of the expression $x^2 – 9$?
Check Solution
Ans: A
The expression $x^2 – 9$ is a difference of squares and can be factored as $(x – 3)(x + 3)$. Therefore, $(x + 3)$ is a factor.
Q.4 What are the roots of the equation $x^2 + 4x + 4 = 0$?
Check Solution
Ans: B
This equation can be factored as $(x + 2)(x + 2) = 0$. Therefore, $x = -2$. Both roots are the same.
Q.5 Solve the quadratic equation $3x^2 – 6x + 2 = 0$ using the quadratic formula.
Check Solution
Ans: A
Applying the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$, where $a = 3$, $b = -6$, and $c = 2$. Therefore, $x = \frac{-(-6) \pm \sqrt{(-6)^2 – 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 – 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.
Next Topic: Discriminant of a Quadratic Equation
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