Question

Form a polynomial whose zeroes are reciprocal of zeroes of 6x² - 7x + 2. *​

Answer 1

Given  :   6x² - 7x + 2

To Find :  a polynomial whose zeroes are reciprocal of zeroes of 6x² - 7x + 2.

Solution:

Method 1 : With out finding zeroes

Let say  α and β  are  zeroes of 6x² - 7x + 2

α + β = -(-7)/6 = 7/6

αβ  = 2/6   = 1/3

1/α  and 1/β  are roots of polynomial

=> 1/α + 1/β  = ( β  + α)/αβ   =  (7/6)/(2/6)  = 7/2

(1/α)  (1/β )   = 1/αβ  =  3

Polynomial = x² - (sum of roots) + product of roots

x²  - (7/2) x  + 3  

multiplying by constant does not change zeroes

=>   2x²  -7x  + 6  is the required polynomial

Method 2 :

Finding zeroes

6x² - 7x + 2 = 0

=> 6x² - 3x - 4x + 2 = 0

=> 3x( 2x - 1) - 2(2x - 1) = 0

=> (3x - 2) (2x - 1) = 0

=>  x= 2/3  or x = 1/2

Reciprocals are

3/2  and  2

( x  - 3/2) (x - 2)   is the required polynomial

(2x - 3)(x - 2) /2

= ( 2x²  - 7x  + 6)/2

multiplying by 2  ( multiplying by constant does not change zeroes )

= 2x²  - 7x  + 6

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Answer 2

Given Equation:-

• 6x² - 7x + 2

To Find:-

•  a polynomial whose zeroes are reciprocal of zeroes of 6x² - 7x + 2.

Solution:-

Finding zeroes

6x² - 7x + 2 = 0

→ 6x² - 3x - 4x + 2 = 0

→ 3x( 2x - 1) - 2(2x - 1) = 0

→ (3x - 2) (2x - 1) = 0

→  x= 2/3  or x = 1/2

Reciprocals of zeroes are, 3/2  and  2

→ ( x  - 3/2) (x - 2)  

→ (2x - 3)(x - 2) /2

multiplying both

→ ( 2x²  - 7x  + 6)/2

multiplying by 2  ( multiplying by constant does not change zeroes )

→ 2x²  - 7x  + 6

hope it helps u ✌