Topics
Linear Equations in Two Variables
- Foundation of Linear Equations in Two Variables
- Simultaneous Linear Equations
- Graph of a Linear Equation in Two Variables
- Methods of Solving Linear Equations in Two Variables > Graphical Method
- Methods of Solving Linear Equations in Two Variables > Determinant method (Cramer’s Rule)
- Equations Reducible to Linear Equations
- Application of Simultaneous Equations
Quadratic Equations
- Concept of Quadratic Equations
- Factorisation Method
- Completing the Square Method
- Quadratic Formula (Shreedharacharya's Rule)
- Nature of Roots of a Quadratic Equation
- Relation Between Zeroes (Roots) and Coefficients of a Quadratic Equation
- Formation of a Quadratic Equation with Given Roots
- Application of Quadratic Equation
Arithmetic Progression
Financial Planning
Probability
Statistics
Introduction
There are several ways to solve a quadratic equation, such as factorisation, using the quadratic formula, and graphing. However, some equations cannot be factorised easily using simple integers. In such cases, completing the square provides a reliable and systematic method.
Definition: Completing the Square
Completing the square is a method of rewriting a quadratic expression so that it contains a perfect square.
For example:
\[ x^{2} + 10x + 25 = (x + 5)^{2} \]
Here, 25 is added to make the expression a perfect square.
General Formula: Completing the Square Method
When an equation is in the standard form \[ x^{2} + bx + c = 0, \] it can be rewritten by adding and subtracting \[ \left(\frac{b}{2}\right)^{2}: \]
\[ x^{2} + bx + \left(\frac{b}{2}\right)^{2} - \left(\frac{b}{2}\right)^{2} + c = 0 \]
This simplifies to the form: \[ \left(x + \frac{b}{2}\right)^{2} = \left(\frac{b}{2}\right)^{2} - c \]
Example 1
Solve: \[ x^{2} + 8x - 48 = 0 \]
Step 1: Shift the constant term.
\[ x^{2} + 8x = 48 \]
Step 2: Add \[ \left(\frac{8}{2}\right)^{2} = 16 \] to both sides.
\[ x^{2} + 8x + 16 = 48 + 16 \]
\[ (x + 4)^{2} = 64 \]
Step 3: Take square roots.
\[ x + 4 = \pm 8 \]
Step 4: Solve.
1. \[ x + 4 = 8 \implies x = 8 - 4 = 4 \]
2. \[ x + 4 = -8 \implies x = -8 - 4 = -12 \]
\[ x = 4 \quad \text{or} \quad x = -12 \]
Answer:
\[ \boxed{x = 4 \text{ or } x = -12} \]
Example 2
Problem: Solve \[ 5x^{2} - 4x - 3 = 0. \]
Step 1: Make the coefficient of \[ x^{2} \] equal to \[ 1 \] by dividing the entire equation by \[ 5: \]
\[ x^{2} - \frac{4}{5}x - \frac{3}{5} = 0 \]
Step 2: Determine the term to complete the square.
Comparing with \[ (x - a)^{2} = x^{2} - 2ax + a^{2}, \]
we get \[ -2ax = -\frac{4}{5}, \] which gives \[ a = \frac{2}{5} \] and \[ k = a^{2} = \frac{4}{25}. \]
Step 3: Add and subtract \[ \frac{4}{25}: \]
\[ x^{2} - \frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - \frac{3}{5} = 0 \]
Step 4: Express the left side as a square:
\[ \left(x - \frac{2}{5}\right)^{2} - \frac{19}{25} = 0 \implies \left(x - \frac{2}{5}\right)^{2} = \left(\frac{\sqrt{19}}{5}\right)^{2} \]
Step 5: Solve for \[ x: \]
\[ x - \frac{2}{5} = \pm \frac{\sqrt{19}}{5} \]
\[ \therefore\; x = \frac{2 \pm \sqrt{19}}{5} \]
Final Answer: The roots of the equation are \[ \frac{2 + \sqrt{19}}{5} \] and \[ \frac{2 - \sqrt{19}}{5}. \]
Key Points: Completing the Square
The completing the square method is used when a quadratic equation cannot be factorised easily.
- First, write the equation in standard form.
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If a ≠ 1, divide the entire equation by a.
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Add and subtract \[\left(\frac{b}{2}\right)^2\] to complete the square.
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Convert the left-hand side into a perfect square.
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Solve the resulting equation to find the roots.
