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Quadratic Equations - Class 10 : Notes


Notes for quadratic equations chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

 

(1) A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is a{x^2} + bx + c, where a, b, c are real number such that a  \ne 0 and x is a real variable.
For Example: {x^2} + 5x + 3, where a = 1,b = 5,c = 3 are real number. So given equation is quadratic polynomial.

(2) If p(x) = a{x^2} + bx + c, a \ne 0 is a quadratic polynomial and \alpha is a real number, then p(\alpha ) = a{\alpha ^2} + b\alpha + c is known as the value of the quadratic polynomial p(\alpha )
For Example: p(\alpha ) = {\alpha ^2} + 5\alpha + 3 in that equation if \alpha=3 then p(\alpha ) = 27. So 27 is a value of quadratic polynomial

(3) A real number \alpha is said to be a zero of quadratic polynomial p(x) = a{x^2} + bx + c, if p(\alpha ) = 0.
For Example: p(x) = {x^2} + 6x + 5
If x =( - 5) then p(x) =0, So -5 is the zero of polynomial.

(4) If p(x) = a{x^2} + bx + c is a quadratic polynomial, then p(x) = 0 i.e., a{x^2} + bx + c = 0, a \ne 0 is called a quadratic equation.
For Example: p(x) = {x^2} - 8x + 16 is a quadratic polynomial, then p(x) = 0 i.e., {x^2} - 8x + 16 = 0, a \ne 0 is called a quadratic equation.

(5) A real number \alpha is said to be a root of the quadratic equation a{x^2} + bx + c = 0.
In other words,   \alpha is a root of a{x^2} + bx + c = 0 if and only if \alpha is a zero of the polynomial p(x) = a{x^2} + bx + c.
For Example: Suppose quadratic equation is 2{x^2} - x - 6 = 0.
If we put x = 2 then p(x)=0, So 2 is a root of that given equation so here \alpha=2 .

(6) If a{x^2} + bx + c = 0, a \ne 0 is factorizable into a product of two linear factors,  then the roots of the quadratic equation a{x^2} + bx + c = 0 can be found by equating each factor to zero.
For Example: The Given equation is 
Now, Solving the above equation using factorization method.


(3x + 1) (3x - 2) =0
 (3x + 1) = 0 or (3x - 2) = 0
3x = -1 or 3x = 2
   or 
Hence,  and   are the two roots of the given equation

 (7) The roots of a quadratic equation can also be found by using the method of completing the square.
For Example:The Given equation is -
                                       
Dividing through out by 2
                                    
Shifting the constant term to the right hand side.
                                   
Adding square of the half of coefficient of x on the both side.
                                   
                                   
                                  
                                  
Taking square root of both sides-
                                  
                                 
                                 
                                 
                                 
Hence x= 3, and x=1/2 are the two root of the given equation


(8) The roots of the quadratic equation a{x^2} + bx + c = 0, a \ne 0 can be found by using the quadratic formula {{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}} , provided that \sqrt {{b^2} - 4ac} \ge 0.
For Example:  the given equation in the form of ,
Where a= √3, b=10 c= 8√3
Therefore, the discriminant- 
D= (10)2 – 4 x √3 x (-8√3)
D= 100 + 96
D= 196
Since, D > 0
Therefore, the roots of the given equation are real and distinct.
The real roots α and β are given by,



For, 


Hence  and  are the two root of the given equation.


(9) Nature of the roots of quadratic equation a{x^2} + bx + c = 0, a \ne 0 depends upon the value of D = {b^2} - 4ac, which is known as the discriminate of the quadratic equation.
For Example: Value of D can be (i) D > 0 , (ii) D = 0 (iii) D < 0.

(10) The quadratic equation a{x^2} + bx + c = 0 , a \ne 0 has:
(i) Two distinct real roots, if D ba-4ac 0 two equal roots i.e. coincident real roots if D = {b^2} - 4ac > 0
For Example:  16x2 = 24x + 1
16x2 – 24x – 1 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 16, b = -24, c = -1
Therefore, the discriminant- D = b2 – 4ac
D= (-24)2 – 4 x 16 x (-1)
D= 576 + 64
D= 640
Since, D > 0
Therefore, the roots of the given equation are real and distinct.
The real roots α and β are given by,





For, 





Hence  and  are the two root of the given equation.


(ii) Two equal roots i.e. coincident real roots, if D = {b^2} - 4ac = 0.
For Example: 2x2 - 2√6x + 3 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 2, b = - 2√6, c = 3
Therefore, the discriminant- D = b2 – 4ac
= (- 2√6)2 – 4 x 2 x 3
= 24 - 24
= 0
Since, D = 0
Therefore, the roots of the given equation are real.
The real and equal roots are given by    and  .

(iii) No real roots, if D = {b^2} - 4ac < 0.
For Example: The given equation is
x2 + x + 2 = 0
The given equation is of the form of ax2 + bx + c = 0, where a = 1, b = 1, c = 2
Therefore, the discriminant
D = b2 – 4ac
D= (1)2 – 4 x 1 x 2
D= 1 - 8
D= -7
Since, D < 0
Therefore, the given equation has not real roots.



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