Understanding Quadratic Equations: Concept, Formula, And Practice Problems

Introduction to Quadratic Equations

Quadratic equations are a crucial part of algebra and are frequently tested in exams like SSC, IBPS, JEE, and various school and competitive exams. A quadratic equation is a polynomial equation of degree 2, which means the highest exponent of the variable (x) is 2. It appears in the form:

\[ ax^2 + bx + c = 0 \]

Where:

• \( a, b, c \) are constants,
• \( a \neq 0 \),
• \( x \) is the variable.

In this post, we will break down the concept of quadratic equations, explore methods to solve them, and provide practice problems to solidify your understanding.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is written as:

\[ ax^2 + bx + c = 0 \]

Where:

• \( a, b, c \) are constants,
• \( a \neq 0 \).

Methods to Solve Quadratic Equations

1. Factoring: If the quadratic equation can be factored, you can break it down into simpler linear expressions.

   Example: Solve \( x^2 - 5x + 6 = 0 \).

   Factor: \((x - 2)(x - 3) = 0\).

   Therefore, \( x = 2 \) and \( x = 3 \).

2. Using the Quadratic Formula: The quadratic formula is a universal method to solve any quadratic equation. It is given by:

   \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

   Where:

   • \( b^2 - 4ac \) is the discriminant. If it's positive, the equation has two real solutions. If it's zero, there's one real solution, and if it's negative, the solutions are complex.

   Example: Solve \( 2x^2 + 4x - 6 = 0 \).

   Here, \( a = 2 \), \( b = 4 \), and \( c = -6 \).

   Using the quadratic formula:

   \[ x = \frac{-4 \pm \sqrt{(4)^2 - 4(2)(-6)}}{2(2)} \]

   \[ x = \frac{-4 \pm \sqrt{16 + 48}}{4} \]

   \[ x = \frac{-4 \pm \sqrt{64}}{4} = \frac{-4 \pm 8}{4} \]

   So, \( x = 1 \) or \( x = -3 \).

3. Completing the Square: This method involves rearranging the equation so that one side becomes a perfect square trinomial.

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Practice Problems With Solutions

Problem 1: Solve the quadratic equation by factoring: \( x^2 + 7x + 10 = 0 \)

Solution: We need to factor the quadratic equation.

\( x^2 + 7x + 10 = 0 \)

Factor the quadratic:

\( (x + 5)(x + 2) = 0 \)

Now, set each factor equal to zero:

\( x + 5 = 0 \quad \text{or} \quad x + 2 = 0 \)

So, \( x = -5 \) or \( x = -2 \).

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Problem 2: Use the quadratic formula to find the roots of the equation: \( 3x^2 - 5x - 2 = 0 \)

Solution: Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 3 \), \( b = -5 \), and \( c = -2 \).

First, calculate the discriminant:

\[ \Delta = b^2 - 4ac = (-5)^2 - 4(3)(-2) = 25 + 24 = 49 \]

Now, apply the quadratic formula:

\[ x = \frac{-(-5) \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7}{6} \]

Thus, we have two solutions:

\[ x = \frac{5 + 7}{6} = 2 \quad \text{or} \quad x = \frac{5 - 7}{6} = -\frac{1}{3} \]

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Problem 3: Solve by completing the square: \( x^2 + 6x + 8 = 0 \)

Solution: Step 1: Move the constant term to the other side of the equation.

\( x^2 + 6x = -8 \)

Step 2: To complete the square, add \( \left( \frac{b}{2} \right)^2 \) to both sides, where \( b = 6 \).

\( \left( \frac{6}{2} \right)^2 = 9 \)

Add 9 to both sides:

\( x^2 + 6x + 9 = -8 + 9 \)

\( (x + 3)^2 = 1 \)

Step 3: Take the square root of both sides:

\( x + 3 = \pm 1 \)

Step 4: Solve for \( x \):

\( x = -3 + 1 = -2 \quad \text{or} \quad x = -3 - 1 = -4 \)

So, \( x = -2 \) or \( x = -4 \).

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Problem 4: Use the quadratic formula to solve: \( x^2 - 4x + 1 = 0 \)

Solution: Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -4 \), and \( c = 1 \).

First, calculate the discriminant:

\[ \Delta = (-4)^2 - 4(1)(1) = 16 - 4 = 12 \]

Now, apply the quadratic formula:

\[ x = \frac{-(-4) \pm \sqrt{12}}{2(1)} = \frac{4 \pm \sqrt{12}}{2} \]

\[ x = 2 \pm \sqrt{3} \] 

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Problem 5: Solve by factoring: \( 2x^2 - 8x + 6 = 0 \)

Solution: First, divide the entire equation by 2 to simplify:

\( x^2 - 4x + 3 = 0 \)

Now, factor the quadratic equation:

\( (x - 3)(x - 1) = 0 \)

Set each factor equal to zero:

\( x - 3 = 0 \quad \text{or} \quad x - 1 = 0 \)

So, \( x = 3 \) or \( x = 1 \).

Conclusion

Quadratic equations are an integral part of algebra and can be solved using different techniques, such as factoring, the quadratic formula, and completing the square. The practice problems provided above will help you test your knowledge and improve your skills in solving these equations. Keep practicing, and soon you'll master this essential concept!