Friday, May 24, 2013

63-04 Basics of Quadratic Equations (Grades 9 to 12) Part-04

Blog-63

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In the Previous Blog we had seen some important concepts of Quadratic Equation.

Some special and critical types of the factors:
Please download the following file and study it very carefully so that you will not find any difficulties while solving quadratic equations. 
Click Here to download the file " Critical-type-of-factors.pdf "
Just go through this downloaded file and be prepared to solve any problem pertaining to these critical factors.

Now we will see important problems of quadratic equations to be solved using perfect square method:

Few steps to be followed to solve the quadratic equation using perfect square method:
              1) Shift the constant term to the right hand side of the equation.
              2) Divide both sides of the equation by the coefficient of  2. 
              3) Find the third term of left hand side to make it as a perfect square.
              4) Use the formula " Third Term = ( 1/2 coefficient of  x ) 2. 
              5) Add this third term so obtained as mentioned in step 4 both sides of the equation.

Please REMEMBER above 5 steps to get your answer quickly and without any mistake. 
a]  Solve the quadratic equation  x 2 - 18 x  + 65 = 0 using perfect square method.

Solution:
      1) Shift 65 to RHS
      2)  2 - 18 x  = - 65                  Here third term = ( 1/2 coefficient of  x ) 2 
                                                                                = ( 1/2 (18) 2
                                                                                = ( 2
                                                                                = 81
           2 - 18 x  + 81 = - 65 + 81
                   (  - 9 ) 2 = 16
                     (  - 9 )  = + 4 or  - 9 )  = - 4 
                                 = 9 + 4 or   =  9 - 4 
                                 = 13 or   =  5
        3)  So the roots of the equation are 5 or 13 so Solution Set = { 5, 13 }  

b]  Solve the quadratic equation  x 2 - 5 x  + 6 = 0 using perfect square method.

Solution:
      1) Shift 6 to RHS
      2)  2 - 5 x  = - 6                  Here third term = ( 1/2 coefficient of  x ) 2 
                                                                                = ( 1/2 (5) 2
                                                                                = ( 5/2 2
                                                                                = 25/4
           2 - 5 x  + 25/4 = - 6 + 25/4
                   (  - 5/2 ) 2 = (25-24)/4
                   (  - 5/2 ) 2 = 1/4
                     (  - 5/2 )  = + 1/2 or  - 5/2 )  = - 1/2 
                                 = 5/2 + 1/2 or   =  5/2 - 1/2 
                                 = 6/2 or   =  4/2
                                 = 3 or   =  2
        3)  So the roots of the equation are 2 or 3 so Solution Set = { 2, 3 }  

c]  Solve the quadratic equation  x 2 - 6 x  + 2 = 0 using perfect square method.

Solution:
      1) Shift 2 to RHS
      2)  2 - 6 x  = - 2                  Here third term = ( 1/2 coefficient of  x ) 2 
                                                                                = ( 1/2 (- 6) 2
                                                                                = ( - 3 2
                                                                                = 9
           2 - 6 x  + 9 = - 2 + 9
                   (  - 3 ) 2 =  7
                     (  - 3 )  = + √ 7 or  - 3 )  = - √ 7 
                                 = 3 + √ 7 or   =  3 √ 7
                                 3 + √ 7 or   =  3 - √ 7
        3)  So the roots of the equation are 3 + √ 7 or 3 - √ 7 so Solution Set = { 3 + √ 7,  3 - √ 7 }  

d]  Solve the quadratic equation  x 2 - 5 x  + 2 = 0 using perfect square method.

Solution:
      1) Shift 2 to RHS
      2)  2 - 5 x  = - 2                  Here third term = ( 1/2 coefficient of  x ) 2 
                                                                                = ( 1/2 (- 5) 2
                                                                                = ( - 5/2 2
                                                                                = 25/4
           2 - 5 x  + 25/4 = - 2 + 25/4
                   (  - 5/2 ) 2 =  (- 8 + 25)/4
                   (  - 5/2 ) 2 =  17/4
                     (  - 5/2 )  = + (√ 17)/2 or  - 5/2 )  = -  (√ 17)/2 
                                 = 5/2 + (√ 17)/2 or   =  5/2 - (17)/2
                                 = (5 + √ 17)/2 or   =  (5 - √ 17)/2
        3)  So the roots are (5 + √ 17)/2 or (5 - √ 17)/2 so Solution Set = { (5 + √ 17)/2,  (5 + √ 17)/2 }  

Please write your opinions about the methods given for these problems. My email id is : anil@7pute.com

Few more problems will be discussed in the next Blog.