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Quadratic Equations - Solve quadratic equations by factorisation in real-root cases

Grade 10CBSE

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

🔑Concepts

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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 a≠0a \neq 0. Visually, this equation represents a parabola; solving the equation means finding where this parabola crosses the horizontal xx-axis.

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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.

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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).

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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 a⋅ca \cdot c. Once found, the middle term bxbx is replaced by these two numbers to allow for factoring by grouping.

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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 (x−α)(x−β)=0(x - \alpha)(x - \beta) = 0, then either x−α=0x - \alpha = 0 or x−β=0x - \beta = 0, leading to the solutions x=αx = \alpha and x=βx = \beta.

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Perfect Square Trinomials are special cases where the quadratic factors into two identical linear factors, such as (x−k)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.

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The Difference of Squares is a shortcut for equations in the form x2−k2=0x^2 - k^2 = 0. This factors directly into (x−k)(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 p⋅q=a⋅cp \cdot q = a \cdot c

Factorised Form: (x−α)(x−β)=0(x - \alpha)(x - \beta) = 0

Difference of Squares: a2−b2=(a−b)(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: a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2

💡Examples

Problem 1:

Solve the quadratic equation x2−7x+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 1⋅12=121 \cdot 12 = 12 and add to −7-7. These numbers are −3-3 and −4-4. \nStep 3: Split the middle term: x2−3x−4x+12=0x^2 - 3x - 4x + 12 = 0. \nStep 4: Group the first two and last two terms: (x2−3x)−(4x−12)=0(x^2 - 3x) - (4x - 12) = 0. \nStep 5: Factor out common terms: x(x−3)−4(x−3)=0x(x - 3) - 4(x - 3) = 0. \nStep 6: Factor out the common binomial: (x−3)(x−4)=0(x - 3)(x - 4) = 0. \nStep 7: Apply the Zero Product Property: x−3=0x - 3 = 0 or x−4=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: 6x2−x−2=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: 6x2−4x+3x−2=06x^2 - 4x + 3x - 2 = 0. \nStep 4: Factor by grouping: 2x(3x−2)+1(3x−2)=02x(3x - 2) + 1(3x - 2) = 0. \nStep 5: Extract the common factor: (3x−2)(2x+1)=0(3x - 2)(2x + 1) = 0. \nStep 6: Solve for xx: 3x−2=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.