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Mathematics 17 Online
OpenStudy (anonymous):

if the sum of the zeroes of the polynomial px^+5x+8p is equal to the product of the zeroes find the value of p...

OpenStudy (apoorvk):

@Parvathysubhash - is that a px^2?

OpenStudy (anonymous):

ya

OpenStudy (maheshmeghwal9):

if a & b are roots/zeroes then;\[a+b=ab\] according to ur question:) then it it; \[\frac{-b}{a}=\frac{c}{a}.\] And ur equation is as\[ax^2+bx+c=0.\]

OpenStudy (apoorvk):

Okay, now.. For a quadratic equation of the standard form\[ax^2 + bx + c =0\]the sum of the roots is (-b/a) and the product of the roots is (c/a). Here your equation is px62 + 5x + 8p = 0, can you compare this with the standard equation and tell me what the 'a' 'b' and 'c' are for this case?

OpenStudy (maheshmeghwal9):

put the values to get ur answer some common sense would also be used:)

OpenStudy (apoorvk):

** i mean "px^2" over there, not "px62".

OpenStudy (anonymous):

@apoorvk ...a =p, b=5 ,c=8p

OpenStudy (apoorvk):

very well! now product of roots = c/a and sum of roots = -b/a. so what would their value be?

OpenStudy (apoorvk):

*values

OpenStudy (anonymous):

-b/a = -5/p, c/a =8p/p

OpenStudy (anonymous):

thx i gt it actually i 4gt to cancel

OpenStudy (anonymous):

p=-5/8??

OpenStudy (apoorvk):

YES!! Right-O!!!!!! :)

OpenStudy (anonymous):

thnq..:D

OpenStudy (apoorvk):

Anytime! :P

OpenStudy (anonymous):

den nw...hav nthr question

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