krit.club logo

Algebra - Quadratic Equations and their Applications

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A quadratic equation is a second-degree polynomial equation in a single variable xx, where the highest power of the variable is 22.

•

The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where a,b,a, b, and cc are real numbers and a≠0a \neq 0.

•

A root or solution of a quadratic equation is a value of xx that makes the equation true.

•

The Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. That is, if (x−p)(x−q)=0(x - p)(x - q) = 0, then x−p=0x - p = 0 or x−q=0x - q = 0.

•

Factorization is the process of breaking down the quadratic expression into a product of two linear binomials, usually by splitting the middle term bxbx into two terms whose coefficients sum to bb and multiply to acac.

•

In real-world applications, quadratic equations often model the area of geometric shapes, projectile motion, or relationships between consecutive integers.

📐Formulae

ax2+bx+c=0ax^2 + bx + c = 0

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

(x+p)(x+q)=x2+(p+q)x+pq(x + p)(x + q) = x^2 + (p + q)x + pq

x2−a2=(x−a)(x+a)x^2 - a^2 = (x - a)(x + a)

💡Examples

Problem 1:

Solve the quadratic equation x2−7x+10=0x^2 - 7x + 10 = 0 by factorization.

Solution:

x2−7x+10=0x^2 - 7x + 10 = 0 x2−5x−2x+10=0x^2 - 5x - 2x + 10 = 0 x(x−5)−2(x−5)=0x(x - 5) - 2(x - 5) = 0 (x−2)(x−5)=0(x - 2)(x - 5) = 0 x−2=0 or x−5=0x - 2 = 0 \text{ or } x - 5 = 0 x=2,x=5x = 2, x = 5

Explanation:

To solve by factoring, find two numbers that multiply to +10+10 and add to −7-7. These numbers are −2-2 and −5-5. Split the middle term and factor by grouping.

Problem 2:

Solve for xx: 4x2−9=04x^2 - 9 = 0.

Solution:

4x2−9=04x^2 - 9 = 0 (2x)2−(3)2=0(2x)^2 - (3)^2 = 0 (2x−3)(2x+3)=0(2x - 3)(2x + 3) = 0 2x−3=0⇒x=322x - 3 = 0 \Rightarrow x = \frac{3}{2} 2x+3=0⇒x=−322x + 3 = 0 \Rightarrow x = -\frac{3}{2}

Explanation:

This is a difference of two squares problem. We use the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) where a=2xa = 2x and b=3b = 3.

Problem 3:

The area of a rectangular garden is 40 m240 \text{ m}^2. The length of the garden is 3 m3 \text{ m} more than its width. Find the dimensions of the garden.

Solution:

Let the width be ww meters. Then length is w+3w + 3 meters. w(w+3)=40w(w + 3) = 40 w2+3w−40=0w^2 + 3w - 40 = 0 (w+8)(w−5)=0(w + 8)(w - 5) = 0 w=−8 or w=5w = -8 \text{ or } w = 5 Since width cannot be negative, w=5 mw = 5 \text{ m}. Length =5+3=8 m= 5 + 3 = 8 \text{ m}.

Explanation:

Set up a quadratic equation based on the area formula Area=Length×WidthArea = Length \times Width. After solving the equation, discard the negative solution as physical lengths must be positive.