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.
No Comments Yet - be the First!