Find discriminant of the following quadratic equation and examine the nature of real roots (if they exist): 7Y2 + 4Y + 5 =0
Answers
The discriminant of a quadratic equation is = b² - 4 ac
Given the equation is :
7 y² + 4 y + 5 = 0
Comparing with a x² + b x + c = 0
a = 7
b = 4
c = 5
Discriminant = b² - 4 ac
= (4)² - 4.7.5
= 16 - 140
= -124
The discriminant is - 124
Since -124 < 0 , hence the equation has no real roots !
Hope it helps
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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 you.
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