Equations and Inequalities: Equations


Variable: The value or an element, which is liable to vary or change called variable. It is commonly represented by a letter.

Constant: The value which remains fixed is called constant. It is commonly represented by a number.

Example: \(2x + 4y = 0\), Here x and y are the variables, 2 and 4 are the constants.

Types of Equations:

(1). Linear Equation: The equation which contains one or more variables with single power is called a linear equation.

Example(1): \(2x + 3 = 4x + 1\), This is the one variable linear equation, here only one 'x' is the variable.

Example(2): \(4x + 5y = 2\), This is the two variable linear equation, here x and y are the two variables.

Quadratic equation: A equation having one variable in the form of \(ax^2 + bx + c = 0\) is called quadratic equation.

Here, x is the variable whereas a, b, and c, are the constants. But \(a \neq 0\), because if \(a = 0\) then the first term of the equation \(ax^2 = 0\), so the equation will be a linear equation.

Example: $$ 2x^2 + 3x + 1 = 0 $$ $$ 2x^2 + 2x + x + 1 = 0 $$ $$ 2x \ (x + 1) + 1 \ (x + 1) = 0 $$ $$ (x + 1) \ (2x + 1) = 0 $$ $$ x = -1, \ and \ x = - \frac{1}{2} $$

Factoring by Inspection Method: This is the most preferable method to solve the quadratic equation, but if it is not possible to factorize the equation by this method, then use the quadratic formula given below in the next paragraph.

Example: $$ 4x^2 + 3x - 1 = 0 $$ $$ 4x^2 + 4x - x - 1 = 0 $$ $$ 4x \ (x + 1) - 1 \ (x + 1) = 0 $$ $$ (x + 1) \ (4x - 1) = 0 $$ $$ x = -1, \ and \ x = \frac{1}{4} $$

Quadratic Formula: If \(ax^2 + bx + c = 0\), then $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
Where, if \(b^2 - 4ac \lt 0\), then there is no real solution.

if \(b^2 - 4ac = 0\), then the equation has one real solution.

if \(b^2 - 4ac \gt 0\), then the equation has two different real solutions.

Example: \(2x^2 + 4x + 2 = 0\)

Solution: Here \(a = 2\), \(b = 4\), and \(c = 2\), so we will find the value of $$ b^2 - 4ac $$ $$ = 4^2 - 4 \times 2 \times 2 $$ $$ = 16 - 16 = 0 $$ Here \(b^2 - 4ac = 0\), so we will get one real solution. $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-4 \pm \sqrt{0}}{2 \times 2} $$ $$ x = \frac{-4}{4} $$ $$ x = -1 $$




Equations and Inequalities

Chapter 1: Equations

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