Quadratic Equations : Quadratic Equations ax^2 + bx +c =0 is a quadratic equation in the standard form.
a,b and c will have real values and a not equal to zero
bor c or both can be zero
x is a variable and the highest power will be two
Examples of quadratic equations : Examples of quadratic equations 2 x^2 + 3x + 5 = 0
5x^2 + 6x = 0
8x^2 = 512
8x^2 + 5x +10 =0
6 x^2 + 12 x = 0
How many values of x can satisfy each equation?
Can you make a guess ? Two values
Solving Factorable Quadratic Equations : Shift all terms to the same side
Factorise the algebraic expression.
Equate each factor to 0.(If the product of two factors equals 0, then either one or both of the factors must be 0.)
Solve each resulting linear equation to arrive at two values for x
Set of two values satisfy the equation and form the solution set Solving Factorable Quadratic Equations Factorisation Method
Solve for x using factorisation method : Solve for x using factorisation method x2 +2x – 35 = 0product: -35 Sum : +2
x2 +7x – 5x – 35 =0
x(x+7) – 5(x+7) =0
(x +7)(x-5) =0
X+7=0 or x-5=0 X= =-7 or x=+5
Solve for x using factorisation method : Solve for x using factorisation method Solve 2 x^2 -19x - 33=0 for x by factoring.
2x ^2 – 22x + 3x– 33 = 0
(Product : -66 Sum : -19)
2x ( x – 11) + 3( x – 11) = 0
( 2x + 3 ) ( x – 11 ) = 0
2x + 3 = 0 , x – 11 = 0
x = – 3/2 , x = 11 X=-3/2 or x=11
Formula Method : Formula Method (b2 - 4ac) is called the discriminant
if it is positive, you will get two solutions
if it is zero, only one solution,
and if it is negative you get two solutions are imaginary [- b +/- Sq. Root(b^2 – 4 ac)]
2 a X=
Solve the equation by using the quadratic formula. 2x2 +3x – 5 = 0a = 2, b = 3 , c = – 5 : Solve the equation by using the quadratic formula. 2x2 +3x – 5 = 0a = 2, b = 3 , c = – 5 x = [- b +/- Sq. Root(b^2 – 4 ac)]
2 a
= [ -3 +/- Sq.Root( 3^2 – 4 (2)(-5)]
2(2)
= [-3 +/-Sq. Root 49]/ 2*2
x = (-3 +/- 7)/4 x = 4/4=1 or -5/2
X^2 + 6x -11 = 0 (a = 1, b = 6, c = -11) : X^2 + 6x -11 = 0 (a = 1, b = 6, c = -11) x = [- b +/- Sq. Root(b^2 – 4 ac)]
2 a
= [ -6 +/- Sq.Root( 6^2 – 4 (1)(-11)]
2(1)
= [-6 +/-Sq. Root 80]/ 2
x = -3 +/- 2Sq.root 5 x = -3 +/- 2Sq.root 5