Question

Y square minus 1 upon 2 Y + 1 upon 16 verify the relation between zeros and its coefficient

Answer 1

Hi ,

Let p(y )= y² - ( 1/2)y + 1/16

To find zeroes make p( y ) = 0

y² - y/2 + 1/16 = 0

16y² - 8y + 1 = 0

( 4y - 1 )² = 0

4y - 1 = 0 or 4y - 1 = 0

y = 1/4 or 1/4

Therefore ,

Two equal zeroes are 1/4 and 1/4


Compare given polynomial with ay² + by + c

a = 16,b = -8 , c = 1 ;

1 ) sum of the zeroes = ( - b / a )

= - ( -8 )/16 = 1/2

= ( 1/4 ) + 1/4 = 1/2

2 ) product of the zeroes = c/a

= 1/16

= ( 1/4 )( 1/4 )

I hope this helps you.

:)