find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficient of the polynomial 1. 7y2 -11/3y -2/3
Answers
= S(2s - 1) - √2(2s - 1)
= (2s - 1)(s - √2)
2s - 1 = 0 ⇒ s = 1/2
s - √2 = 0 ⇒ s = √2
Therefore, Zeroes of the polynomial are 1/2 and √2.
If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then
α +β = − b/a
αβ = c/a.
Therefore, sum of the roots is (1 + 2√2) / 2 i.e. (1/2 + √2)
Product of the roots is √2 / 2 = 1/√2.
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 you
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