please guys answer this question
Answers
Solution:-
x
3
+3x
2
+4x−11=0
Here,
A=1,B=3,C=4,D=−11
Given that a,b,c are the roots of the above equation.
Therefore,
Sum of roots =
A
−B
⇒a+b+c=
1
−3
=−3
Sum of the product of roots =
A
C
⇒ab+bc+ca=
1
4
=4
Product of roots =
A
−D
⇒abc=
1
−(−11)
=11
x
3
+rx
2
+sx+t=0
Here,
A=1,B=r,C=s,D=t
Given that (a+b),(b+c),(c+a) are the roots of the above equation.
Therefore,
Sum of the roots =
A
−B
⇒(a+b)+(b+c)+(c+a)=
1
−r
=−r
Sum of the product of roots =
A
C
⇒(a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)=
1
s
=s
Product of the roots =
A
−D
⇒(a+b)(b+c)(c+a)=
1
−t
=−t
Now, solving for 't', we have
t=−[(a+b)(b+c)(c+a)]
⇒t=−[ab+ac+b
2
+bc)(c+a]
⇒t=−[a
2
b+a
2
c+2abc+ab
2
+b
2
c+ac
2
+bc
2
]
⇒t=−[a
2
b+a
2
c+abc+abc+ab
2
+b
2
c+ac
2
+bc
2
+abc−abc]
⇒t=−[a(ab+ac+bc)+b(ac+ab+bc)+c(ac+bc+ab)−abc]
⇒t=−[(ab+bc+ca)(a+b+c)−abc]
From eq
n
(1),(2)&(3), we have
t=−[4×(−3)−11]
t=−(−12−11)=23
Hence the value of 't' is 23.