If the roots of r- 14r?+ 56r - 64 = 0 are in G.P then the middle root is
Answers
Solution :-
Given Cubic equation is
r³-14r²+56r-64 = 0
Given that
The roots of the given equation are in the G.P.
Let they be a/r , a , ar
Where, a = first term and
r = common ratio
On comparing this with the standard cubic equation ax³+bx²+cx+d = 0
a = 1
b = -14
c = 56
d = -64
We know that
Sum of the roots = -b/a
=> (a/r)+(a)+(ar) = -(-14)/1
=> (a+ar+ar²)/r = 14 --------(1)
and
Product of the roots = -d/a
=> (a/r)(a)(ar) = -(-64)/1
=> (a×a×ar)/r = 64
=> a×a×a = 64
=> a³ = 64
=> a³ = 4³
=> a = 4
If a = 4 then equation (1) becomes
=> (4+4r+4r²)/r = 14
=> 4+4r+4r² = 14r
=> 4+4r+4r²-14r = 0
=> 4r²-10r+4 = 0
=> 2(2r²-5r+2) = 0
=> 2r²-5r+2 = 0
=> 2r²-4r-r+2 = 0
=> 2r(r-2)-1(r-2) = 0
=> (r-2)(r-1) = 0
=> r-2 = 0 or r-1 = 0
=> r = 2 or r = 1
If a = 4 and r = 2 then ar = 4(2) = 8
and a/r = 4/2 = 2
Therefore, a/r = 2 , a = 4 , ar = 8
If a = 4 and r = 1 then ar = (4)(1) = 4
and a/r = 4/1 = 4
Therefore, a/r = 4 , a = 4 ,ar = 4
We notice that The middle root = 4
Alternative Method :-
Given Cubic equation is
r³-14r²+56r-64 = 0
It can be written as
=> r²-8r²-6r²+48r+8r-64 = 0
=> [r²(r-8)-6r(r-8)+8(r-8)] = 0
=> (r²-6r+8)(r-8) = 0
=> [(r²-2r-4r+8)(r-8)] = 0
=> [(r-2)(r-4)](r-8) = 0
=> (r-2)(r-4)(r-8) = 0
=> r-2 = 0 or r-4 = 0 or r-8 = 0
=> r = 2 or r = 4 or r = 8
The roots of the given equation are 2,4,8
first term = 2
Common ratio = 4/2 = 8/4 = 2
They are in the G.P.
Therefore, The middle root = 4
Answer :-
The middle root of the given equation is 4
Used formulae:-
♦ The standard cubic equation is ax³+bx²+cx+d = 0
♦ Sum of the roots = -b/a
♦ Product of the roots = -d/a
♦ Sum of the product of the two roots taken at a time = c/a