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Quadratic Equations - Solution of Quadratic Equations by Factorisation

Grade 10CBSE

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

🔑Concepts

A Quadratic Equation is an algebraic equation of the second degree, typically written in the standard form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,a, b, and cc are real numbers and a0a \neq 0. Visually, this equation represents a parabola; solving the equation means finding where this parabola crosses the horizontal xx-axis.

The Roots of a quadratic equation are the values of xx that satisfy the equation. If we substitute a root into the equation, the result is zero. On a graph, these roots are the xx-intercepts of the quadratic function.

The Factorisation Method involves breaking down the quadratic expression ax2+bx+cax^2 + bx + c into a product of two linear factors, such as (mx+n)(px+q)=0(mx + n)(px + q) = 0. This is based on the idea that a large area (the quadratic) can be represented as the product of its length and width (the linear factors).

Splitting the Middle Term is the primary technique for factorisation. To factor ax2+bx+cax^2 + bx + c, you must find two numbers whose sum equals bb and whose product equals aca \cdot c. Once found, the middle term bxbx is replaced by these two numbers to allow for factoring by grouping.

The Zero Product Property states that if the product of two numbers or expressions is zero, then at least one of them must be zero. Mathematically, if (xα)(xβ)=0(x - \alpha)(x - \beta) = 0, then either xα=0x - \alpha = 0 or xβ=0x - \beta = 0, leading to the solutions x=αx = \alpha and x=βx = \beta.

Perfect Square Trinomials are special cases where the quadratic factors into two identical linear factors, such as (xk)2=0(x - k)^2 = 0. Visually, this means the vertex of the parabola sits exactly on the xx-axis, and we say the equation has two equal real roots.

The Difference of Squares is a shortcut for equations in the form x2k2=0x^2 - k^2 = 0. This factors directly into (xk)(x+k)=0(x - k)(x + k) = 0, representing a parabola symmetric about the yy-axis with roots at kk and k-k.

📐Formulae

Standard Form: ax2+bx+c=0ax^2 + bx + c = 0

Splitting Criteria: Find p,qp, q such that p+q=bp + q = b and pq=acp \cdot q = a \cdot c

Factorised Form: (xα)(xβ)=0(x - \alpha)(x - \beta) = 0

Difference of Squares: a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b)

Square Identity 1: a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

Square Identity 2: a22ab+b2=(ab)2a^2 - 2ab + b^2 = (a - b)^2

💡Examples

Problem 1:

Solve the quadratic equation x27x+12=0x^2 - 7x + 12 = 0 by factorisation.

Solution:

Step 1: Identify a=1,b=7,c=12a = 1, b = -7, c = 12. \nStep 2: Find two numbers that multiply to 112=121 \cdot 12 = 12 and add to 7-7. These numbers are 3-3 and 4-4. \nStep 3: Split the middle term: x23x4x+12=0x^2 - 3x - 4x + 12 = 0. \nStep 4: Group the first two and last two terms: (x23x)(4x12)=0(x^2 - 3x) - (4x - 12) = 0. \nStep 5: Factor out common terms: x(x3)4(x3)=0x(x - 3) - 4(x - 3) = 0. \nStep 6: Factor out the common binomial: (x3)(x4)=0(x - 3)(x - 4) = 0. \nStep 7: Apply the Zero Product Property: x3=0x - 3 = 0 or x4=0x - 4 = 0. \nFinal Answer: x=3x = 3 and x=4x = 4.

Explanation:

We use the splitting the middle term method. Since the product is positive (12) and the sum is negative (-7), both factors must be negative.

Problem 2:

Solve for xx: 6x2x2=06x^2 - x - 2 = 0.

Solution:

Step 1: Identify a=6,b=1,c=2a = 6, b = -1, c = -2. \nStep 2: Find two numbers with product 6(2)=126 \cdot (-2) = -12 and sum 1-1. These numbers are 4-4 and 33. \nStep 3: Split the middle term: 6x24x+3x2=06x^2 - 4x + 3x - 2 = 0. \nStep 4: Factor by grouping: 2x(3x2)+1(3x2)=02x(3x - 2) + 1(3x - 2) = 0. \nStep 5: Extract the common factor: (3x2)(2x+1)=0(3x - 2)(2x + 1) = 0. \nStep 6: Solve for xx: 3x2=0    x=233x - 2 = 0 \implies x = \frac{2}{3} or 2x+1=0    x=122x + 1 = 0 \implies x = -\frac{1}{2}. \nFinal Answer: x=23,12x = \frac{2}{3}, -\frac{1}{2}.

Explanation:

In this case, aa is not 1, so we must multiply aa and cc to find the target product (-12). We then find factors of -12 that sum to -1 and group the terms to solve.