Question

if y=3 is a root of a quadratic equation 7y2-(k+1)y+3=0 find the value of k

Answer 1

Answer:

Step-by-step explanation:Hi ,

Let p( y ) = 7y² - 11y/3 - 2/3 ,

To find the zeroes , we have to take

p ( y ) = 0

7y² - 11y/3 - 2/3 = 0

Multiply each term with ' 3 ' we get

21y² - 11y - 2 = 0

21y² - 14y + 3y - 2 = 0

7y ( 3y - 2 ) + 1( 3y - 2 ) = 0

( 3y - 2 ) ( 7y + 1 ) = 0

Therefore ,

3y - 2 = 0 or 7y + 1 = 0

3y = 2 or 7y = -1

y = 2/3 or y = ( -1/7 )

Therefore ,

Required two zeroes of p( y ) are

m = 2/3 , n = -1/7

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Compare p( y ) with ax² + bx + c , we

get

a = 7 , b = -11/3 , c = 2/3 ,

1 ) sum of the zeroes = -b / a

= - ( -11/3 )/ 7

= 11/21 ----( 1 )

m + n = 2/3 - 1/7

= ( 14 - 3 ) / 21

= 11/21 ---( 2 )

( 1 ) = ( 2 )

2 ) product of the zeroes = c/a

= ( -2/3 ) /7

= - 2/21----( 3 )

mn = (2/3 ) ( - 1/7 )

=- 2/21 ----( 4 )

(3 ) = (4 )

I hope this helps u

Answer 2

Answer:

Value of k = 21

Step-by-step explanation:

Given that y=3 is a root of 7y²-(k+1)y+3=0

so it satisfied the given equation

we put y=3 in the given equation

7y²-(k+1)y+3=0

7×3²-(k+1)×3+3=0

7×9-(k+1)×3+3=0

63-(k+1)×3+3=0

66-(k+1)×3=0

66=(k+1)×3

66÷3=k+1

22=k+1

22-1=k

k=21