Some special and critical types of factors:
Please download the following file and study it very carefully so that you will not find any difficulties while solving quadratic equations. Download the following file by clicking on it.
Click here to download the critical-type-of-factors.pdf file.
Just go through this downloaded file and be prepared to solve any problem pertaining to these critical factors.
Now we will see a 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 2 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 }
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 2 is 1 and the sign of the constant term 143 is " + ".
2) So, x 2 - 24 x + 143 = 0 Here 143 is odd so 2, 4, 6, and 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. The 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 the same as the digit in 10th place so 11 is the
the factor of 143 so here factors 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 }
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 2 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 }
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 2 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 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 }
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 2 is 1 and the sign of the constant term 18 is " + ".
2) So, x 2 + 5 √3 x + 18 = 0 Here the factors of 1 & 18 with addition as 5 √3 are
1 x 18
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 2 √3 & 3 √3
6 x 15
x 2 + 2 √3 x + 3 √3 x + 18 = 0
so we have to take 2 √3 & 3 √3
6 x 15
x 2 + 2 √3 x + 3 √3 x + 18 = 0
x ( x + 2 √3 ) + 3 √3 ( x + 2 √3 ) = 0
( x + 2 √3 ) ( x + 3 √3 ) = 0
( x + 2 √3 ) = 0 or ( x + 3 √3 ) = 0
x = - 2 √3 or x = - 3 √3
3) So the roots of the equation are - 2 √3 or - 3 √3 so Solution Set = { - 2 √3 - 3 √3 }
( x + 2 √3 ) = 0 or ( x + 3 √3 ) = 0
x = - 2 √3 or x = - 3 √3
3) So the roots of the equation are - 2 √3 or - 3 √3 so Solution Set = { - 2 √3 - 3 √3 }
A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.
Click here for the next basics.
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