An equation which is of the form ax2+bx+c=0 where a,b,c are real numbers and a≠ 0 is a Quadratic Equation
For Example: 3x2+4x+5=0, 7x2-5x-3=0 and 6-x2+100=0 are Quadratic Equations.
Hence we conclude that any equation which is of the form p(x) = 0 , Where p(x) is a polynomial of degree 2 is a Quadratic Equation.
We see that these Quadratic Equations arise in several situations in the world around us and also been used in different fields of Mathematics.
To understand this we shall look at some examples
Rohan and Joel together have 80 pebbles.Each of them lost 10 pebbles and the product of the number of the pebbles they now is 540. We would like to find out the number of pebbles they had initially.
Let the number of pebbles Rohan had be taken as x
Then the number of pebbles Joel had will be 80-x
The number of pebbles left with Rohan when he lost 10 pebbles = x-10
The number of pebbles left with Joel when he lost 10 pebbles = 80-x-10 = 70-x
Hence their product
= (x-10) (70-x)
So, -x2+80x-700 = 540 [ Since their product is given as 540 ]
i.e., -x2 +80x -160 =0
i.e., x2-80x+160 = 0
Therefore the number of marbles Rohan had satisfies the Quadratic Equation
x2-80x+160 = 0
This gives the solution for the above situation Mathematically.
The area of the rectangular plot is 270 m2. The length of the plot ( in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. Represent this situation in the form of Quadratic Equation.
Let the breadth of the rectangular plot be x
Given that the length is one more than twice its breadth
Hence we have
We know that area of the rectangle = length * breadth
(2x+1) x= 270
2x2+x-270=0 is the solution for the given problem which is in the form of a Quadratic Equation.
Check whether the following are Quadratic Equations
- (x-3)2+4= 2x+3Solution:Consider L.H.S = (x-3)2+4 = x2-6x+9+4= x2-6x+13
Therefore, (x-3)2+4 = 2x+3 can be rewritten as
i.e., x2-8x+10 = 0
The above equation is of the form ax2+bx+c= 0
Therefore, the above equation is a Quadratic Equation.
- x(x+1)+8= (x+2) (x-2)Solution:Consider L.H.S x(x+1)+8= x2+x+8
And R.H.S (x+2) (x-2) = x2-4
i.e., x2+x+8= x2-4
The above equation is not of the form ax2+bx+c = 0
Therefore, the above equation is not a Quadratic Equation.