70-11 Basics of Quadratic Equations

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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.

Some Important Problems related to Quadratic Equations:

Anil Satpute

69-10 Basics of Quadratic Equations

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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.

Some Important Problems related to Quadratic Equations:

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

Click here for the next basics.

Anil Satpute

68-09 Basics of Quadratic Equations

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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.

Some Important Problems related to Quadratic Equations:

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

Click here for the next basics.

Anil Satpute

67-08 Basics of Quadratic Equations

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Some Important Problems related to Quadratic Equations:

 
a) If the sum of the root of the quadratic equation [1/(x + m)] + [1/(x + n)] = 1/p is 0, then show that the product of the roots is -[(m ² + n ²)/2]

Now by using the above method, we can easily solve the following examples.

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

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Anil Satpute

66-07 Basics of Quadratic Equations

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Some Important Problems related to Quadratic Equations:

a) If one root of the quadratic equation k x ² - 5 x + 2 = 0 is 4 times the other, then find " k "
Solution:

b) If the roots of the quadratic equation  x ² +  m x + n = 0 differ by 3, then prove that m ² =  4 n + 9
Solution:

c) If one root of the quadratic equation p x ² +  x + p = 0 is the square of the other, then find  prove that 1 + p q ( p + q - 3 ) = 0
Solution:

Hence Proved

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

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Anil Satpute

65-06 Basics of Quadratic Equations

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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 and 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 in Quadratic equations 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 - 3 α β (α + β)

                                   so ,     63      =  (3)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 

                       x ² - (α  +  β)  x + (α β) = 0 

                       x ² - (3)  x + (- 4) = 0 

                       x ² - 3 x - 4 = 0 

          4)  Answer: So the required quadratic equation will be x ² - 3 x - 4 = 0.

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

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Anil Satpute

64-05 Basics of Quadratic Equations

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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 x ² + b x  = - c 

      2) Dividing the equation by " a " throughout 

          So,   x ² + (b / a) x  = - (c / a) 

               Here, third term = ( 1/2 coefficient of  x ) ² 

                                         = ( 1/2 (b / a) ) ²

                                         = (b / 2 a) ²

                                         = b 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  x ² - 4 x - 21 = 0 with a x ² + b x + c = 0 we get,

            a = 1, b = - 4, c = - 21

       2) Here Discriminant

                                                =  (- 4 )² - 4 (1) ( - 21) 

                                                =  16 + 84       

                                                =  100

A few more problems on Quadratic Equations related to factors will be discussed in the next Blog.

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Anil Satpute

63-04 Basics of Quadratic Equations

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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 x ². 
              3) Find the third term of the left-hand side to make it 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 on both sides of the equation.

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

Solution:

      1) Shift 65 to RHS

      2)  x ² - 18 x  = - 65                  Here third term = ( 1/2 coefficient of  x ) ² 

                                                                                = ( 1/2 (18) ) ²

                                                                                = ( 9 ) ²

                                                                                = 81

           x ² - 18 x  + 81 = - 65 + 81

                   ( x  - 9 ) ² = 16

                     ( x  - 9 )  = + 4 or ( x  - 9 )  = - 4 

                                x  = 9 + 4 or  x  =  9 - 4 

                                x  = 13 or  x  =  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)  x ² - 5 x  = - 6                  Here third term = ( 1/2 coefficient of  x ) ² 

                                                                                = ( 1/2 (5) ) ²

                                                                                = ( 5/2 ) ²

                                                                                = 25/4

           x ² - 5 x  + 25/4 = - 6 + 25/4

                   ( x  - 5/2 ) ² = (25-24)/4

                   ( x  - 5/2 ) ² = 1/4

                     ( x  - 5/2 )  = + 1/2 or ( x  - 5/2 )  = - 1/2 

                                x  = 5/2 + 1/2 or  x  =  5/2 - 1/2 

                                x  = 6/2 or  x  =  4/2

                                x  = 3 or  x  =  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)  x ² - 6 x  = - 2                  Here third term = ( 1/2 coefficient of  x ) ² 

                                                                                = ( 1/2 (- 6) ) ²

                                                                                = ( - 3 ) ²

                                                                                = 9

           x ² - 6 x  + 9 = - 2 + 9

                   ( x  - 3 ) ² =  7

                     ( x  - 3 )  = + √ 7 or ( x  - 3 )  = - √ 7 

                                x  = 3 + √ 7 or  x  =  3 - √ 7

                                x  = 3 + √ 7 or  x  =  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)  x ² - 5 x  = - 2                  Here third term = ( 1/2 coefficient of  x ) ² 

                                                                                = ( 1/2 (- 5) ) ²

                                                                                = ( - 5/2 ) ²

                                                                                = 25/4

           x ² - 5 x  + 25/4 = - 2 + 25/4

                   ( x  - 5/2 ) ² =  (- 8 + 25)/4

                   ( x  - 5/2 ) ² =  17/4

                     ( x  - 5/2 )  = + (√ 17)/2 or ( x  - 5/2 )  = -  (√ 17)/2 

                                x  = 5/2 + (√ 17)/2 or  x  =  5/2 - (√17)/2

                                x  = (5 + √ 17)/2 or  x  =  (5 - √ 17)/2

     3)  So the roots are (5 + √ 17)/2 or (5 - √ 17)/2 so Solution Set = {(5 + √ 17)/2,  (5 + √ 17)/2 }  

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A few more problems on Quadratic Equations will be discussed in the next Blog.

Click here for the next basics.

Anil Satpute