## Thursday, May 23, 2013

### 62-03 Basics of Quadratic Equations (Grades 9 to 12) Part-03

Blog-62

Dear Students,

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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 few more important problems of quadratic equations:

d]  Solve the quadratic equation  x 2 - 18 x  + 65 = 0 using factorization method.

Solution:
1) The coefficient of x  is 1 and the sign of the constant term  65 is " + ".
2)  So,           x 2 - 18 x  + 65 = 0                  Here the factors of 1 & 65 are
5 x 13
as we want the sum as 18
so we have to take 5 & 13
5 x 13
x 2 - 5 x - 13 x  + 65 = 0
x ( x  -  5 )  -  13 ( x  -  5 ) = 0
( x  -  5 )  ( x  -  13 ) = 0
( x  -  5 )  =  0  or  ( x  -  13 ) = 0
x  =  5  or   x  =  13
x  =  5  or  x  =  13
3)  So the roots of the equation are 5 or 13 so Solution Set = { 5, 13 }

e]  Solve the quadratic equation  x 2 - 24 x  + 143 = 0 using factorization method.

Solution:
1) The coefficient of x  is 1 and the sign of the constant term  143 is " + ".
2)  So,           2 - 24 x  + 143 = 0                  Here 143 is odd so 2, 4, 6, 8 are not the factors of
143. Similarly 1 + 4 + 3 = 8, which is not divisible by 3
& 9 so 3 & 9 are also not the factors of 143. Unit digit
is not 5 or 0 so 5 and 10 are not the factors of 143.
Sum of digit at unit place & 100th place is 1 + 3 = 4
which is same as the digit at 10th place so 11 is the
factor of 143  so here the factors of 1 & 143 will be
11 x 13
x 2 - 11 x - 13 x  + 143 = 0
x ( x  -  11 )  -  13 ( x  -  11 ) = 0
( x  -  11 )  ( x  -  13 ) = 0
( x  -  11 )  =  0  or  ( x  -  13 ) = 0
x  =  11  or   x  =  13
x  =  5  or  x  =  13
3)  So the roots of the equation are 11 or 13 so Solution Set = { 11, 13 }

f]  Solve the quadratic equation  5 x 2 + 56 x  + 11 = 0 using factorization method.

Solution:
1) The coefficient of x  is 5 and the sign of the constant term  11 is " + ".
2)  So,          5 x 2 + 56 x  + 11 = 0                  Here the factors of 5 & 11 with addition as 56 are
55 x 1
as we want the sum as 56
so we have to take 55 & 1
55 x 13
5  x 2 + 55 x  +  x  + 11 = 0
5 x ( x  +  11 )  +  ( x  +  11 ) = 0
( 5 x  +  1 )  ( x  +  11 ) = 0
( 5 x  +  1 )  =  0  or  ( x  +  11 ) = 0
5 x  =  - 1  or  x  =  - 13
x  =  - 1/5  or  x  =  - 11
3)  So the roots of the equation are - 1/5 or - 11 so Solution Set = { - 1/5, - 11 }

g]  Solve the quadratic equation  2 x 2 + 21 x  + 45 = 0 using factorization method.

Solution:
1) The coefficient of x  is 2 and the sign of the constant term  45 is " + ".
2)  So,          2 x 2 + 21 x  + 45 = 0                  Here the factors of 2 & 45 with addition as 21 are
2 x 45
2 x 5 x 9
2 x 5 x 3 x 3
(2 x 3) x (5 x 3)
6 x 15
as we want the sum as 21
so we have to take 6 & 15
6 x 15
2 x 2 + 6 x  + 15 x + 45 = 0
2 x ( x  +  3 )  + 15 ( x  +  3 ) = 0
( 2 x  +  15 )  ( x  +  3 ) = 0
( 2 x  +  15 )  =  0  or  ( x  +  3 ) = 0
2 x  =  - 15  or  x  =  - 3
x  =  - 15/2  or  x  =  - 3
3)  So the roots of the equation are - 15/2 or - 3 so Solution Set = { - 3, - 15/2 }
h]  Solve the quadratic equation   x 2 + 5 3 x  + 18 = 0 using factorization method.

Solution:
1) The coefficient of x  is 1 and the sign of the constant term  18 is " + ".
2)  So,          2 + 5 3 x  + 18 = 0                  Here the factors of 1 & 18 with addition as 5 are
1 x 18
3 x 2 x 3
(2 3) x (3 3)
as we want the sum as 5 3
so we have to take 3 & 3
6 x 15
x 2 + 3 x  + 3 x + 18 = 0
x ( x  +  3 )  + 3 ( x  +  3 ) = 0
( x  +  3 )  ( x  +  3 ) = 0
( x  +  3 )  =  0  or  ( x  +  3 ) = 0
x  =  - 3  or  x  =  - 3
3)  So the roots of the equation are - 3 or - 3 so Solution Set = { - 3  - 3 }
Few more problems will be discussed in the next Blog.

Please write your opinions about the methods given for these problems.

Anil Satpute