if one of the zeroes of the quadratic polynomial 2x^2 + 8x + 2x + 1 is seven times the other then find the value of k.
pls answer fast
i will mrk u brnlst
monalisaborah48:
Where is the k in the question??
your question is wrong
Edit it
Answers
Answered by
8
Answer: have a look at the picture
Step-by-step explanation:
Attachments:
products of zeros = c/a
Srry mate...sometimes I'm really too careless
I have edited it...really sorry for the mistakes
good bro
*sis
Now I came to know where does my marks goes away....silly mistakes(though this was a big one!)....thanks for informing bro
well tried bro
and thnxx
Wels sis...I'm also sis not bro(^_^)
oh sorry my lovely siso
Answered by
11
Question:- if one of the zeroes of the quadratic polynomial 2x^2 + 8x + 2k + 1 is seven times the other then find the value of k.
Answer: 5/4
Step-by-step explanation:
Given,
p(x) = 2x² + 8x + 2k + 1
if the zeroes of following is α and β ,then according to question,
β = 7α.
Comparing this polynomial with standard form of quadratic polynomial
ax² + bx + c ,
a = 2
b = 8
c = 2k + 1
We know that,
Sum of zeroes = -b/a
α + β = -8/2
α + 7α = -4
8α = -4
α = -4/8
α = -1/2
β = 7 × -1/2 = -7/2
Now,
Product of zeroes = c/a
αβ = (2k+1)/2
(-1/2) × (-7/2) = (2k+1)/2
7/4 = (2k+1)/2
7/2 = 2k + 1
7 = 4k + 2
4k = 7 - 2
4k = 5
k = 5/4
thnxx bro
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