plz answer ques 10 .... plzzz i will be grateful
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Aaryadeep:
just wait bro for a picture... well are you in class 9 ?
yss
i m also in class 9 :-)
gudd
can u answer me one more question
yes sure is it this one or any other
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I just tried.... hope this helps..
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thankss
welcome
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hi see this pic for your answer
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wait a minute I have some work.....
hey bro if you put x=√-2 then there will be two zero and the question say that it has one zero
yes...but we cannot write -√2.... so in complex no. i²= -1 therefore = (√2i)^3 - 3(√2i)^2 + 2√2i - 6= 2√2i^3 - 3×2×i^2 + 2√2i - 6 = -2√2i + 6 + 2√2i - 6 = 6-6 = 0 hence proved...
no bro...it has one zero only..you can try this steps...
Supposing you mean:
p(x) = x^3 - 3x^2 + 2x - 6
This is a cubic equation, so it has three zeros by its nature.
use the rational roots theorem to find the root at x = 3.
Use the factor theorem to convert this root into a factor:
(x - 3)
Divide by the known factor:
(x^3 - 3x^2 + 2x - 6) / (x - 3) = x^2 + 2
Use the zero product principle:
x^2 + 2 = 0
Subtract 2:
x^2 = -2
Take the square root:
x = ± i √2 where i^2 = -1
So what's been shown is that p(x) has only one REAL zero.
p(x) = x^3 - 3x^2 + 2x - 6
This is a cubic equation, so it has three zeros by its nature.
use the rational roots theorem to find the root at x = 3.
Use the factor theorem to convert this root into a factor:
(x - 3)
Divide by the known factor:
(x^3 - 3x^2 + 2x - 6) / (x - 3) = x^2 + 2
Use the zero product principle:
x^2 + 2 = 0
Subtract 2:
x^2 = -2
Take the square root:
x = ± i √2 where i^2 = -1
So what's been shown is that p(x) has only one REAL zero.
& another is complex root
& for your kind information manu1121 we cannot write -√2 in this ex.
thanks for ur help afshar ... thnku
it's ok bro
hmm i got it
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