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The world of quadratic equations

An equation of the form ax²+bx+c=0 is called a quadratic equation, where a,b,c are known values (i.e. constants), a is non-zero and x is the unknown value (i.e. a variable). 
For ex: 5x²+7x+3=0, 4x²+2=0, 3x²+8x+4=8, are all quadratic equations.

Since the highest power of a quadratic equation is 2, a quadratic equation can have at the most two unique solutions (which are also called the roots of the equation). Thus, for a given quadratic equation, x can have at the most 2 unique values.

NOTE: An equation of the type ax²+bx+d=e where a,b,c,d are known constants can be reduced t the form of a quadratic equation i.e.

ax²+bx+d=e                       

=> ax²+bx+c=0     where c = d-e

Deriving a formula to find the solution of a quadratic equation:
Consider a quadratic equation ax²+bx+c=0                   - (1)

To solve this equation we have to convert the terms containing x into a perfect square, i.e. we have to convert ax²+bx into a perfect square as these are the terms containing x.

To make ax² into a perfect square we shall multiply it with a. Hence, we multiply equation (1) with 'a' on both sides.

a²x² + abx + ac = 0        - (2)

To complete the square we require another perfect square term on the LHS, so that we can form a term of the form (a+b)² =a² + b² + 2ab on the LHS for the terms containing x. After having a closer look at the LHS we can see that b² could be the candidate to be added.

So we add b² to both sides of equation (2). Thus we have

a²x² + abx + ac + b² = b²        - (2)

Now we multiply the above equation by the smallest non-unity perfect square i.e. 4 to get

4a²x² + 4abx + 4ac + 4b² = 4b²

=>    (2ax)² + 2(2abx) + 4ac + b² + 3b² = 4b²

=>    (2ax)² + 2(2ax)(b) + b² + 4ac + 3b² = 4b²

=>    (2ax+b)² + 4ac + 3b² = 4b²                              

=>    (2ax+b)²  = b² - 4ac                                   

=>    2ax+b  = +\/b² - 4ac                                  

=>    2ax  = -b + \/b² - 4ac                                 

=>    x  = (-b + \/b² - 4ac )/(2a)                         - (3)

This is the  solution of a quadratic equation. As we can clearly see, the existence of a square root on the RHS indicates that x may have 2 unique values, until and unless b² - 4ac=0. 

Let x₁ and x₂ be the 2 unique solutions i.e. roots of the quadratic equation. Thus we have,
x₁  = (-b + \/b² - 4ac )/(2a) and x₂  = (-b - \/b² - 4ac )/(2a)

SUM OF THE ROOTS
Now let us find the Sum of these roots

Thus  x₁+x₂ = (-b + \/b² - 4ac )/(2a) + (-b - \/b² - 4ac )/(2a)

          x₁+x₂ = (-b + \/b² - 4ac  -b - \/b² - 4ac )/(2a)

          x₁+x₂ = (-b-b)/(2a)

          x₁+x₂ = (-2b)/(2a)

          x₁+x₂ = -b/a

Thus the sum of the roots of any quadratic equation is always -b/a.

PRODUCT OF THE ROOTS
Now let us find the Product of these roots

Thus  x₁x₂ = [(-b + \/b² - 4ac )/(2a) ][ (-b - \/b² - 4ac )/(2a)]

          x₁x₂ = [(-b + \/b² - 4ac)( -b - \/b² - 4ac )]/(4a²)

          x₁x₂ = [b² + b\/b² - 4ac - b\/b² - 4ac - b² + 4ac]/(4a²)

          x₁x₂ = [b² - b² + 4ac]/(4a²)

          x₁x₂ = (4ac)/(4a²)

          x₁x₂ = c/a

Thus the product of the roots of any quadratic equation is always c/a.

-by Gurudev
MADE IN INDIA
gurudevp@vsnl.net
On 07 December 2002

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