Question

If α and β are zeroes of the polynomial 6x2 – 7x –3, then form a quadratic polynomial whose zeroes are 2α and 2β

Answer 1

Sol : We have ,

Sum of roots = - ( coefficient of x ) / ( coefficient of x² )

α + ß = - ( - 7 ) / 6

α + ß = 7 / 6.

Now ,

Product of roots = constant term / coefficient of x²

αß = ( - 3 ) / 6 

αß = - 1 / 2.

Now  a quadratic equation whose zeroes are 2α and 2ß.

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

= x² - ( 2α + 2ß ) x + ( 2α * 2ß )

= x² - ( α + ß ) 2 x + 4 αß

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

= x² - ( 7 / 6 ) 2 x + 4 ( - 1/ 2 )

= x² -  ( 7x / 3  ) - 2

   3 x² - 7 x - 6
= ---------------------- = 0
           3

= 3 x² - 7 x - 6 .Ans.


                          __________________________________

Answer 2

When a and b are zeroes of 6x² - 7x - 3

a + b = 7/6

ab = -3/6 = -1/2

So, Any equation having 2a and 2b as roots can be written as

(x - 2a) ( x - 2b) = 0

= x² - 2( a + b)x + 4ab = 0

= x² - 2 × 7/6 x + 4 × (-1/2) = 0

= x² -7/3 x - 2 = 0

multiply by 3

3x² - 7x - 6 = 0