Question

solveeeeeeeeeeeeeeeeeeeeeeeeeeee
if possible solve in ur copy

Answer 1

                          =======================

Given,

Quadratic equation = x² - 2x + 3.

Here,

Coefficient of x² = 1

Coefficient of x = -2

Constant term = 3

α and ß are its zeroes.

We know the relationship between zeroes and coefficient of a quadratic equation,

⇒ Sum of zeroes = - Coefficient of x ÷ Coefficient of x²

⇒ α + ß = - ( -2 ) ÷ 1

∴ α + ß = 2

Now,

⇒ Product of zeroes = Constant term ÷ Coefficient of x²

⇒ αß = 3 ÷ 1

∴ αß = 3.

For 1st question,

⇒ Sum of zeroes = α + 2 + ß + 2 

                             = α + ß + 4

                             = 2 + 4 = 6.

Product of zeroes = ( α + 2 ) ( ß + 2 )

                              = αß + 2α + 2ß + 4 

                             = αß + 4 + 2 ( α + ß )

By substituting the values of αß and ( α + ß ).

                           = 3 + 4 + 2 ( 2 )

                          = 3 + 4 + 4 = 11.

The general form of a quadratic equation :

⇒ x² - ( Sum of zeroes ) x + Product of zeroes = 0

⇒  x² - 6x + 11 = 0

For 2nd question,

Sum of zeroes = { ( α - 1 ) / ( α + 1 ) } + { ( ß - 1 ) / ( ß + 1 ) }


      ( α - 1 ) ( ß + 1 ) + ( ß - 1 ) ( α + 1 )
= ---------------------------------------------------------
                ( α + 1 ) ( ß + 1 )

    αß + α- ß - 1 + αß + ß - α - 1
= -----------------------------------------------
           αß + α + ß + 1

                2αß - 2
= --------------------------------------
         αß + α + ß + 1

By substituting the values of αß and ( α + ß ),

            2 × 3 - 2
=  ------------------------------------
            3 + 2 + 1

              6 - 2
= ---------------------------
                6

= 4/6 = 2/3.

Product of zeroes = { ( α - 1 ) / ( α + 1 ) } { ( ß - 1 ) / ( ß + 1 ) }

          ( α - 1 ) ( ß - 1 )
= --------------------------------------
        ( α + 1 ) ( ß + 1 )

      αß - α - ß + 1
= ---------------------------------
      αß + α + ß + 1

    αß + 1 - ( α + ß )
= ---------------------------
    αß + 1 + ( α + ß )

By substituting the values of αß and ( α + ß ),

           3 + 1 - 2
= --------------------------------
           3 + 1 + 2
 
            4 - 2
= -------------------------
               6

= 2/6 = 1/3.

The general form of a quadratic equation is :

⇒ x² - ( Sum of zeroes )x + Product of zeroes = o

⇒ x² - ( 2/3 )x + ( 1/3 ) = 0

        3x² - 2x + 1
⇒ ------------------------- = 0
               3

⇒ ( 3x² - 2x + 1 ) = 0 × 3

⇒ 3x² - 2x + 1 = 0.

The required quadratic equation is 3x²- 2x + 1.

Sorry Bro, I didn't solve it in note-book.

Answer 2

Hey mate!

Here's the answer of (1)....

Kindly refer to the given attachment.

Hope it helps you!!