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Quadratic equations

An equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0 is called a quadratic equation.

Examples:

• x² – 8x + 16 = 0  is a quadratic equation in variable ‘x’.
  
• p² –(3/2)p + 8 = 0  is a quadratic equation in variable ‘p’.
  
• a + 1/a = 3/2, when simplified, takes the form 2a² –3a + 2 = 0.
  So, the given equation is a quadratic equation in variable ‘a’.
  
• (m + g)y² – √ 8y + 16√2 = 0  is a quadratic equation in variable ‘y’.
  

Solution of a quadratic equation: We know that ax² + bx + c = 0 is a quadratic equation. For any real number ‘ p ‘ if ap² + bp + c = 0 , then p is called solution or root of the quadratic equation.

Method of solving a quadratic equation: There are two methods to solve a quadratic equation: (i) Factorization method (ii) Quadratic formula.

Zero product rule : Let a and b be any two real numbers.
Then, ab = 0     a = 0 or b = 0.   This is called zero product rule.

Solving a Quadratic equation by completing the square :

The quadratic equation is ax² + bx + c = 0 where a , b, c R and a ≠ 0

Multiply both sides by 4a ≠ 0, we have

                4a(ax² + bx + c)   =   4a(0)

            4a(ax² + bx + c)   =   0

            4a²x² + 4abx + 4ac   =   0

          Adding b² to both sides, we get

               4a²x² + 4abx + 4ac + b²  =   0 + b²

            4a²x² + 4abx + 4ac + b²  =   b²

               Transpose 4ac to the otherside we get,

               4a²x² + 4abx + b²  =   b² – 4ac

            [2ax]² + 4abx + b²  =   b² – 4ac

            [(2ax)² + 2(2ax)(b) + b²]   =   b² – 4ac

            [2ax + b]²  =   b² – 4ac

          This is called quadratic formula

Note:
In the quadratic formula (b² – 4ac) is called discriminant of the standard quadratic equation ax² + bx + c = 0 and it is denoted by D.

Examples:

1) Find the discriminant for k² + 12k + 32 = 0 ?

Sol :

        Given equation is k ² + 12k + 32 = 0.

        By comparing with standard equation ax² + bx + c = 0 we get,

        a = 1 , b = 12 , c = 32

        Now we will calculate the discriminant D = b² –4ac

            D = 12 ² – 4 * 1 * 32 = 144 – 128 = 16.

        The discriminant for the given equation k ² + 12k + 32 = 0 is 16.

2) Find the discriminant for x(x + m – n) = mn ?

Sol :

        Given equation is x(x + m – n) = mn.

            x(x + m – n) – mn = 0.

            x² + (m – n)x – mn = 0

        By comparing with standard equation ax² + bx + c = 0 we get,

        a = 1 , b = (m – n) , c = – mn

        We know that the discriminant D = b² – 4ac

            D = (m – n)² – 4 * (1) * (–(mn))

            D = (m – n)² + 4mn

            D = (m + n)²………     ( since (a – b)² + 4ab = (a + b)²)

        The discriminant for the given equation x(x + m – n) = mn is (m + n)²
.

Quadratic inequalities:
If ax ² + bx + c is a quadratic expression, then
                             ax ² + bx + c > 0
                             ax ² + bx + c ≥ 0
                             ax ² + bx + c < 0
                             ax ² + bx + c ≤ 0
are called quadratic inequalities.

        From above line we can conclude that the common portion is x > 7.  

        The thick black line represents the x > 7 and x > – 3

                 The solution set for (x – 7)(x + 3) > 0   is { x : x > 7}

        Now Take x – 7 < 0 , x + 3 < 0           x < 7 , x < – 3

        Take a number line , represent x < 7 and x < – 3.

        From above line we can conclude that the common portion is x < –3.  

        The thick black line represents the x < 7 , x < –3

                 The solution set for (x – 7)(x + 3) < 0   is { x : x < – 3}

        We have ab     ≥     0             ab > 0 , ab = 0

       Hence (x – 7)(x +3)     ≥     0

            (x – 7)(x +3) > 0 , (x – 7)(x +3) = 0

            (x – 7)(x +3)   =  0

            x   =   7,  –3.

        At x = 7 , –3 also the inequality is satisfying.

            The solution set for x² – 4x – 21   ≥   0 is { x : x ≤ – 3 or x ≥ 7}

Ex2: Solve the inequality 2k² + 3k < 2 ?

Sol :
        Given inequality 2k² + 3k < 2

            2k² + 3k – 2 < 0

            2k² + 4k – k – 2 < 0

            2k(k + 2) – 1 (k + 2) < 0

            (2k – 1)(k + 2) < 0

        We know that ab < 0
                         a > 0 , b < 0     OR     a < 0 , b > 0

        Now, consider (2k – 1 )(k + 2) < 0

                         2k – 1 > 0 , k + 2 < 0     or     2k – 1 < 0 , k + 2 > 0.

                         2k > 1 , k < – 2     or     2k < 1 , k >–2

        From above line it is clear that there is no common portion for ‘k’.