Quadratic Equation in Algebra

Review :

Practice Examples 1  ,  Practice Examples 2

 Quadratic Equation with one variable

The standard form of writing Quadratic equation is  ax2 + bx + c = 0 ,  where a, b and c are real numbers and a ≠0 and x is a variable.

Whenever we square a variable there will be two possible outcomes of it; one is positive and another is negative

If  x2  = 9  then x = 3 or  x = -3

Similarly when x2 = 16 then x = 4 or  x = – 4

Ex.1       8x2 -10x + 3 = 0

Solution   

To solve the quadratic equation we factor it and these are the step we should follow to factor the equation

  • Multiply the constant  with x2  e. a = 8 and constant at place of ‘c’ ( ax2 + bx + c = 0  ) i.e. c = 4
  • 8 x 4  = 24
  • Now factor ‘24’ such that by adding or subtracting the factors you will get middle term i.e. b = -10
  • So, here the factors of 24  will be  -6  and  – 4 by which we get multiple ; 6 x 4 = 24 and  addition  of  (-6)  + (-4) = -10

Factor the equation by taking multiples of middle term ‘10x’ as 6x and 4x

         8x2 – 6x – 4x +3 = 0

2x (4x – 3) – 1(4x – 3) = 0

(4x – 3)  (2x – 1) = 0

(4x-3) = 0 or (2x – 1) = 0

x = 3/4 or x = 1/2

Ex. 2     10m2 – 11m – 6 = 0

Solution;   Factor the equation

10 m2 – 15 m + 4 m – 6 = 0

5m (2m – 3) + 2 (2m – 3) = 0

(2m – 3) or (5m +2) = 0

m = 3/2 or m = -2/5

Ex. 3   Convert into Quadratic Equation

(m – 5) (m + 3) = 0

Solution;

m2 + 3m – 5m – 15 = 0

m2  – 2m  -15  = 0

Ex. 4   Find the solution of   (x-3) (x +5) = 0

Solution;       (x-3) (x + 5) = 0

(x-3) = 0 or (x+5) = 0

x= 3 or x = -5

Ex. 5     Find the solution of   (2m+1) (m-8) = 0

Solution;                (2m+1) (m-8) = 0

(2m +1) = 0   or    (m – 8) = 0

m = -1/2   or   m = 8

In general we solve quadratic equation by factorization  but if the equation is not easily factored that time we find its roots  using following formula;

q2

We will get two real numbers  of ‘x’ after using this formula but;

When  b2-4ac  ≤  0 then we can solve like this –

1.   If  b2-4ac  =  0 then the roots of   ‘x’ will be :

x = -b/2a    and we will have only one root of the equation.

2.   If  b2-4ac  <  0 that time value of √b²-4ac   will not be real  and equation will not have real roots.

Ex. 6     Solve the Equation    5x2 -2x – 16 = 0

Solution;    Here solving the equation with factorization is not possible we will go with roots or ‘x’

a = 5, b = -2,  c = -16

q4

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