Find the zeroes - : y2 +3/2root 5 y - 5
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using
X=b+_√b^2-4ac
here b=3/2√5
a=1 And c=-5
so, now
y= [-3/2√5+_√(3/2√5)^2-4(1)(-5)]/2
y=[-3/2√5+_√125/4]/2
y=[-3/2√5+_5√5/2]/2
therefore on solving roots of of y will
-2√5 &√5/2
X=b+_√b^2-4ac
here b=3/2√5
a=1 And c=-5
so, now
y= [-3/2√5+_√(3/2√5)^2-4(1)(-5)]/2
y=[-3/2√5+_√125/4]/2
y=[-3/2√5+_5√5/2]/2
therefore on solving roots of of y will
-2√5 &√5/2
sakshi169:
i had uploaded the soln
ya i came to knw which one
I need it through factorization method not by shri dharacharya method
okk .....i will try in that way n ans u
wait
sorry i had written the roots wrongly.....sry its -2√5 nd +√5/2
it can be possible by factorization also
write 3/2√5 as 2√5-√5/2.....nd solve by taking common.....u will get ur roots
Oh..Thanks
have u got ur soln
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