answer this plzz..with process..
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ask if any doubt..........
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clearly two roots are 1 and -2 let the third root be p
sum of product of roots taken two at a time is given by c/a in a cubic equation so
(1×(-2)) + (-2p) + (1p) = (-1/2)
-2 -p = -1/2
p= -3/2
products of root = -d/a
1×(-2)×(-3/2)= -(-n/2)
3=n/2
n=6
sum of roots = -b/a
1+ (-2) + (-3/2) = -m/2
-5/2= -m/2
m=5
m^2 - n^2 = 5^2 - 6^2
= 25-36
= -11
option 1
hope it helps
sum of product of roots taken two at a time is given by c/a in a cubic equation so
(1×(-2)) + (-2p) + (1p) = (-1/2)
-2 -p = -1/2
p= -3/2
products of root = -d/a
1×(-2)×(-3/2)= -(-n/2)
3=n/2
n=6
sum of roots = -b/a
1+ (-2) + (-3/2) = -m/2
-5/2= -m/2
m=5
m^2 - n^2 = 5^2 - 6^2
= 25-36
= -11
option 1
hope it helps
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