please solve que 17 and 18
Attachments:
Answers
Answered by
1
17) f(x)=3x²-5x-2
[let me write alpha as 'a' and beta as 'b']
for the zeros a and b f(x)=0
=>3x²-5x-2=0
so, a+b = 5/3
and ab = -2/3
(i)a+b=5/3
=>(a+b)²=25/9
=>a²+2ab+b²=25/9
=>a²+b²-4/3=25/9
=>a²+b²= (25/9)+(4/3)
=>a²+b²=37/9
(ii)a³+b³=(a+b)³-3ab(a+b)
=>a³+b³=(5/3)³-3(-2/3)(5/3)
=>a³+b³=125/27+10/3=215/27
(iii)a²/b+b²/a=(a³+b³)/ab
=>a²/b+b²/a=(215/27)/(-2/3)=-215/18
18.p(x)=6x²+x-1
for zeroes a and b p(x)=0
=>6x²+x-1=0
a+b=-1/6
ab=-1/6
(i)a³b+ab³=ab(a²+b²)
=>a³b+ab³=ab((a+b)²-2ab)=-1/6(1/36+1/3)=-13/216
(ii)a/b+b/a+2(1/a+1/b)+3ab
=a²+b²/ab+2(a+b/ab)-1/2
=-6((a+b)²-2ab)+2-1/2
=-13/6+2-1/2
=(-13+9)/6
=-2/3
[let me write alpha as 'a' and beta as 'b']
for the zeros a and b f(x)=0
=>3x²-5x-2=0
so, a+b = 5/3
and ab = -2/3
(i)a+b=5/3
=>(a+b)²=25/9
=>a²+2ab+b²=25/9
=>a²+b²-4/3=25/9
=>a²+b²= (25/9)+(4/3)
=>a²+b²=37/9
(ii)a³+b³=(a+b)³-3ab(a+b)
=>a³+b³=(5/3)³-3(-2/3)(5/3)
=>a³+b³=125/27+10/3=215/27
(iii)a²/b+b²/a=(a³+b³)/ab
=>a²/b+b²/a=(215/27)/(-2/3)=-215/18
18.p(x)=6x²+x-1
for zeroes a and b p(x)=0
=>6x²+x-1=0
a+b=-1/6
ab=-1/6
(i)a³b+ab³=ab(a²+b²)
=>a³b+ab³=ab((a+b)²-2ab)=-1/6(1/36+1/3)=-13/216
(ii)a/b+b/a+2(1/a+1/b)+3ab
=a²+b²/ab+2(a+b/ab)-1/2
=-6((a+b)²-2ab)+2-1/2
=-13/6+2-1/2
=(-13+9)/6
=-2/3
sweetynagpal:
kya ye shi h jse maine kia
ha
fir answer que glt aaya
tumhara anwser kya aya
6.5/2
to fir arithmetic me galati hogi. formula sahi huya to bass you can do any of the maths with same concept. caio
ek kam kro mrko follow kro mai abhi yhi solution upload krungi mistake bta dena and unfollow kr dena
DM mai message karo then. :/
nhi kr skte show nhi hota
:/ guess the problem remains. anyways gotta go. caio
Similar questions