solve the equation 2x^4 - 15x^3 + 35x^2 - 30x + 8 = 0 , whose roots are in G.P.
Answers
we have to solve the equation 2x⁴ - 15x³ + 35x² - 30x + 8 = 0 , whose roots are in G.P
solution : let a/r³ , a/r , ar , a³ are four terms in GP which are also roots of given biquadratic equation .
sum of all roots = -(-15)/2 = 15/2
⇒a/r³ + a/r + ar + ar³ = 15/2 ...(1)
product of all roots = 8/2 = 4
⇒a/r³ × a/r × ar × ar³ = 4
⇒a⁴ = 4
⇒a = ±√2
now sum of products of three roots = -(-30)/2 = 15
⇒a/r³ × a/r × ar + a/r × ar × ar³ + ar × ar³ × a/r³ + ar³ × a/r³ × a/r = 15
⇒a³[1/r³ + r³ + r + 1/r ] = 15
⇒(2√2)[(1 + r⁴)/r³ + (r⁴ + 1)/r ] = 15
⇒(1 + r⁴)(1 + r²)/r³ = 15/2√2
if we assume r = √2
we get, (1 + 4)(1 + 2)/2√2 = 15/2√2
so, r = √2 is correct
also r = 1/√2 is correct.
so, r = √2 , 1/√2
we take, r = √2 , a = √2
so, first term, a/r³ = (√2)/(2√2) = 1/2
2nd term , a/r = (√2)/√2 = 1
3rd term , ar = √2 √2 = 2
4th term , ar³ = √2 (√2)³ = 4
therefore roots are 1/2, 1 , 2 and 4
Answer:
To solve the equation
2x4−15x3+35x2−30x+8=0
We consider ,the solutions are
α,β,γ,δ
So from theory of equation ,we have
∑α=152....(1)
∑αβ=352...(2)
∑αβγ=15...(3)
αβγδ=4....(4)
Condition that
αβ=γδ
So equation 1,2,3,4 can be written as
α+β+γ+δ=152….5
(α+β)(γ+δ)+αβ+γδ=352….6
αβ(γ+δ)+γδ(α+β)=15…..7
αβγδ=4…8
From (8) we get
(αβ)2=4⇒αβ=±2
If we take αβ=−2
Then equation 5 does not hold
So we take αβ=2
Now from (6) we have
(α+β)(γ+δ)+2+2=352
(α+β)(γ+δ)=272
Again from (5) we get
[(α+β)+(γ+δ)]2
=[(α+β)−(γ+δ)]2+4(α+β)(γ+δ)
[(α+β)−(γ+δ)]
=(152)2–4.272−−−−−−−−−√
=225–2164−−−−−√
=94−−√
=±32….(9)
( We take +ve sign)
Now ,from (1) and (9) we have
α+β=92….(10)
And
γ+δ=3….(11)
Again,f from (10)
(α−β)=(α+β)2–4αβ−−−−−−−−−−−√
=814−4.2−−−−−−−√
=±72….(12)
We take +ve sign
So solving (10) and (12) we get
α=4,β=12
Again ,from (11) we get
(γ−δ)
=(γ+δ)2–4.γδ−−−−−−−−−−−√
=9–4.2−−−−−√
=±1……(13)
We take +ve sign
So,from (11) and (13) we get
γ=2,δ=1
Hence the required solutions are
x=1,12,2,4