Quadratic Equation



                   Quadratic Equation

A quadratic equation in the variable X is an equation of the form
where a,b and c are real numbers, a≠0 2x2+2x+5=0 is an quadratic equation, similarly 6x2+x+2=0, and

4x2-5x+2=0 are also quadratic equation.

Any equation in the form of p(x)=0 is a polynomial of degree 2,  is a quadratic equation. but when we write the terms of p(x) in descending order of their degree, then we get the standard form of the equation. That is, ax2+bx+c=0, a≠0, is called the standard form of quadratic equation.

Example:- (x-2)2+1=2x-3

Solution :-  

L.H.S=(x-2)2+1=x2-4x+4+1
                              
                          [(a-b)2= a2-2ab+b2]
          
           = x2-4x+5

Therefore, (x-2)2+1=2x-3 can be written as
                  
                     x2-4x+5=2x-3
                   
                     x2-4x+5-2x+3=0

i.e,                x2-6x+8=0

it is form ax2+bx+c=0,
therefore, the given equation is a quadratic equation.
  













x
x

Comments

Post a Comment

Popular posts from this blog

NCERT-class-10th - Quadratic equation - Exercise 4.4

Image
Sum No.1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them. (i) 2x 2 -3x+5=0  (ii) 3x 2 -4 √ 3x +4=0 (iii) 2x 2 -6x+3=0      Ans. (i) 2x 2 -3x+5=0  Comparing this equation with general equation ax 2 +bx+c=0, We get a=2, b=-3 and c=5 Discriminant = b 2 -4ac=(-3) 2 -4*2*5   =9-40=-31 Discriminant is less than 0 which means equation has no real roots.   (ii) 3x 2 -4 √ 3x +4=0 Comparing this equation with general equation ax 2 +bx+c=0, We get a=3, b=-4 √ 3 and c=4 Discriminant = b 2 -4ac=(-4 √ 3) 2 -4*3*4 =48-48=0 Discriminant is equal to 0 which means equation has equal real roots. Applying quadratic   x= [-b 士 √b 2 -4ac]/ 2a to find roots ⇒ x= [ 4 √ 3 士 √0 ]/ 2*3  =( 2 √ 3)  /3 Because, equation has two equal roots, it means  x= (2 √ 3)/3 , (2 √ 3)/3   (iii) 2x 2 -6x+3=0   Comparing this equation with general equation ax 2 +bx+c=0, We get a=2, b=-6 and c=3 Discriminant=b 2 -4ac=(-6) 2 -4*2*3  =
Read more

Class 10th-Quadratic Equation- Exercise 4.3.Full notes

Image
1. Find the roots of the following quadratic equations if they exist by the method of completing square. (i) 2x 2 -7x+3=0 (ii) 2x 2 +x-4=0 (iii)4x 2 +4 √ 3x+3=0 (iv) 2x 2 +x+4=0  Ans. (i) 2x 2 -7x+3=0 First we divide equation by 2 to make co-efficient of  x 2  equal to 1, x 2 -(7/2)x+3/2=0 Divide middle term of the equation by 2 x , we get (7/2)x*(1/2x)=7/4  Add and subtract square of 7/4  from the equation  x 2 -(7/2)x+3/2=0 x 2 -(7/2)x+3/2+(7/4) 2 -(7/4) 2 =0 x 2 +(7/4) 2 -7/2 x+3/2+(-7/4) 2 =0 {(a-b) 2 =a 2 +b 2 -2ab} (x-7/4) 2 -(24-49)/16=0 (x-7/4) 2 =(49-24)/16 Taking Square root on both sides, x-7/4= 士 5/4 X=5/4+7/4=12/4=3  or  x=-5/4+7/4=2/4=1/2 Therefore, x=1/2 or 3   (ii) 2x 2 +x-4=0 First we divide equation by 2 to make co-efficient of  x 2  equal to 1, x 2 +x/2-2=0 Divide middle term of the equation by 2 x , we get x/2*1/2x=1/4 Add and subtract square of 1/4  from the equation  x 2 +x/2-2=0 x 2 -x/2-2+(1/4) 2 -(1/4) 2 =0
Read more

Example 3 Step by Step explanation class 10th Quadratic equation

Image
Example :- find the roots of the equation 2x 2 -5x+3 by factorisation. Solution :- Step 1:- firstly split the middle term -5x as -2x-3x                                                                      [because (-2x)*(-3x)]  Step 2:- Now After splitting middle term, we can write the given equation as                            2x 2 -2x-3x+3  =2x(x-1)-3(x-1)=0 Step 3:- now   2x(x-1)-3(x-1)=0                          (2x-3)(x-1)=0                         2x-3=0 or x-1=0   Step 4:- So, the value of x for which 2x 2 -5x+3=0 are the same                   i.e, either 2x-3=0 or x-1=0 Step 5:- Now, 2x-3=0 g ives x=3/2                  And    x-1=0 gives   x=1 So   x= 3/2   and x= 1 is the solution of the equation.   In other word, 1 and 3/2 are the roots of the equation 2x 2 -5x+3 =0     Example:-
Read more