Wednesday, May 22, 2013

02 Basics of Quadratic Equations (Grades 9 to 12) Part-02

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Dear Students,

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

Now we can study few more problems based on roots of the quadratic equation.

c] If one root of the quadratic equation 7 2 - k x - 8 = 0 is 2, find the value of k.
Solution:
      1) As x = 2 is the root of the quadratic equation 2 - k x - 8 = 0, it will satisfy the equation.
      2) So, put x = 2 in  2 - k x - 8 = 0, we get
                      2 - k x - 8 = 0
                (2) 2 - k (2) - 8 = 0
                  (4)  -  2 k  - 8 = 0
                      28  -  2 k  - 8 = 0
                           20  -  2 k  = 0
                                     20  =  2 k
                                     10  =  k
    3) So, the value of k is 10.

To solve the quadratic equation:
         
       1) Using factorization Method:

              a) Solve the quadratic equation 2 - 5 x + 6 = 0

Note: Before solving the quadratic equation, few important concepts need to be understood. 
        1) The roots of the quadratic equation is known as the Solution Set of the quadratic equation, and they are written in the curly brackets.
        2) While solving the quadratic equation by factorization method, first see the sign of the constant term. If the sign is " + " then we need to find a group of two factors of constant term and the coefficient of   2 in such a way that their addition will be the coefficient of x.
Please study this step very carefully.so that you will understand the basics of factorization of quadratic equation or expression.

See the above problem.
           2 - 5 x + 6 = 0, here coefficient of 2 is 1 and the constant term is 6 with " + " sign, so, we need to get a group of two factors of 6 x 1 in such a way that their sum will be (- 5). Here the factors of 6 are 1 and 6 or 2 and 3. Here sum of the factors in the first group is 1 + 6 = 7 which is not 5 so ignore this group. Now see the second group with factors 2 & 3, their sum is 2 + 3 = 5. so we need to replace (- 5) x by (- 2) x and (- 3) x. (See the following step)
           2 - 2 x  -3 x  +  6 = 0
          x (x - 2) - 3 (x - 2) = 0
                   (x - 2) (x - 3) = 0
                   (x - 2)  =  0 or  (x - 3) = 0
                   x  = 2 or  x = 3
                  so, 2, 3 are the roots of the quadratic equation   x 2 - 5 x + 6 = 0. So here Solution Set of this equation is { 2, 3 }.

Now we will solve few more examples of quadratic equation using factorization method.

b]  Solve the quadratic equation 6 x 2 -  x  - 2 = 0 using factorization method.

Solution:
      1) The coefficient of x  is 6 and the sign of the constant term  2 is " - ".
      2)  So,          6 x 2 -  x  - 2 = 0                  Here the factors of 6 & 2 are 
                                                                                          6 x 2
                                                                                          2 x 3 x 2  as we want the difference as 1
                                                                                                   so we have to take (2 x 2) & 3
                                                                                          4 x 3
                            6 x 2 - 4 x + 3 x  - 2 = 0
              2 x ( 3 x  -  2 )  +  ( 3 x  -  2 ) = 0
                        ( 3 x  -  2 )  ( 2 x  +  1 ) = 0
                        ( 3 x  -  2 )  =  0  or  ( 2 x  +  1 ) = 0
                        3 x  =  2  or  2 x  =  - 1
                           x  =  2/3  or  x  =  - 1/2
       3)  So the roots of the equation are 2/3 or - 1/2 so Solution Set = { 2/3, - 1/2 }  


c]  Solve the quadratic equation 4 x 2 - 23 x  + 15 = 0 using factorization method.

Solution:
      1) The coefficient of x  is 4 and the sign of the constant term  15 is " + ".
      2)  So,          4 x 2 - 23 x  + 15 = 0                  Here the factors of 4 & 15 are 
                                                                                4 x 15
                                                                                2 x 2 x 3 x 5  as we want the sum as 23
                                                                                so we have to take (2 x 2 x 5) & 3
                                                                                20 x 3
                            4 x 2 - 20 x - 3 x  + 15 = 0
                      4 x ( x  -  5 )  -  3 ( x  -  5 ) = 0
                        ( x  -  5 )  ( 4 x  -  3 ) = 0
                        ( x  -  5 )  =  0  or  ( 4 x  -  3 ) = 0
                         x  =  5  or  4 x  =  3
                           x  =  5  or  x  =  3/4
       3)  So the roots of the equation are 5 or  3/4 so Solution Set = { 5, 3/4 }  

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.

Few more problems on Quadratic Equation related with factors will be discussed in the next Blog.

Anil Satpute