Find the value of p if sum of zeroes of given polynomial is one third of their product. x2-(2p+1)x+(3p+7)
Answers
Answered by
6
Hey... I think this can be ur answer dear!!
Sum of zeroes =-b/a = - (-(2p+1)/1
=2p+1
Now, product =c/a=3p+7/1
According to the question,
Sum =1/3 product
3(2p+1)=3p+7
6p+3=3p+7
3p=4
p=4/3........
Sum of zeroes =-b/a = - (-(2p+1)/1
=2p+1
Now, product =c/a=3p+7/1
According to the question,
Sum =1/3 product
3(2p+1)=3p+7
6p+3=3p+7
3p=4
p=4/3........
Answered by
4
Given polynomial
x² - (2p+1)x + (3p+7) = 0
By comparing it with ax²+bx+c = 0,we get
a = 1 , b = -(2p+1) , c = 3p+7
Sum of zeroes = -b/a
= -[-(2p+1)]
= 2p+1
Product of zeroes = c/a
= 3p+7
As per problem,
Sum of zeroes = ⅓ × product of zeroes
2p+1 = ⅓ × 3p+7
3(2p+1) = 3p+7
6p+3 = 3p+7
6p - 3p = 7 - 3
3p = 4
=> p = 4/3
x² - (2p+1)x + (3p+7) = 0
By comparing it with ax²+bx+c = 0,we get
a = 1 , b = -(2p+1) , c = 3p+7
Sum of zeroes = -b/a
= -[-(2p+1)]
= 2p+1
Product of zeroes = c/a
= 3p+7
As per problem,
Sum of zeroes = ⅓ × product of zeroes
2p+1 = ⅓ × 3p+7
3(2p+1) = 3p+7
6p+3 = 3p+7
6p - 3p = 7 - 3
3p = 4
=> p = 4/3
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