## Friday, May 31, 2013

Now we will see a few quadratic equations using the formula method.
a) Solve the quadratic equation  5 x 2 - 2 x - 9 = 0.
Solution:
1) Comparing  5 x 2 - 2 x - 9 = 0 with a x 2 + b x + c = 0 we get,
a = 5, b = - 2, c = - 9

b) Solve the quadratic equation  3 x 2 - 4 x - 3 = 0.
Solution:
1) Comparing  3 x 2 - 4 x - 3 = 0  with a x 2 + b x + c = 0 we get,
a = 3, b = - 4, c = - 3

Now we will see some rules about the roots of the quadratic equation.
1) Sum of the roots (α, β) of the quadratic equation a x 2 + b x + c = 0,
α  +  β  =  - b / a  and   α β  =  c / a
α  +  β  =  - Coefficient of x / Coefficient of x 2  and   α β  =  Constant / Coefficient of x 2.
2) If α, β are the roots of the quadratic equation then the quadratic equation will be
x 2 - (Sum of the roots)  x + (Product of the roots) = 0,
x 2 - (α  +  β)  x + (α β) = 0,

Now we will see some important problems on Quadratic equation and the roots of the Quadratic Equations

a) If the sum of the roots of the quadratic equation is 3 and the sum of their cubes is 63, then find the equation.
Solution:
1)  Here, α + β =  3 and  α 3 + β 3 =  63
2)  We know that , α 3 + β 3 =  (α + β)- 3 α β (α + β)
so ,     63      =  (3)- 3 α β (3)
63      =  27 - 9 α β       Dividing by 9 through out we get
7       =  3  - α β
α β   =  3  - 7
α β   =  - 4
3) We know that  the quadratic equation will be
2 - (α  +  β)  x + (α β) = 0
2 - (3)  x + (- 4) = 0
2 - 3 x - 4 = 0
4)  Answer: So the required quadratic equation will be 2 - 3 x - 4 = 0.
Few more problems will be discussed in the next Blog.

## Thursday, May 30, 2013

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

Solution:
1) Shift c to RHS we get  a 2 + b x  = - c
2) Dividing the equation by " a " through out
So,   2 + (b / a) x  = - (c / a)                  Here third term = ( 1/2 coefficient of  x ) 2
= ( 1/2 (b / a) 2
= (b / 2 a2
= 2/ 4 a 2

Now we will see a few quadratic equations using the formula method.
a) Solve the quadratic equation  x 2 - 4 x - 21 = 0.
Solution:
1) Comparing  2 - 4 x - 21 = 0 with a x 2 + b x + c = 0 we get,
a = 1, b = - 4, c = - 21
=  (- 4 )2 - 4 (1) ( - 21)
=  16 + 84
=  100

Few more problems on Quadratic Equations will be discussed in the next Blog

Anil Satpute

## Friday, May 24, 2013

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

Few steps to be followed to solve the quadratic equation using the 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 the 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.

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 }

Few more problems will be discussed in the next Blog.

## Thursday, May 23, 2013

Some special and critical types of the factors:
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.