If x²– 5x + 1 = 0, then
x¹⁰+1/ x⁵
has the value:
(A) 2524 (B) 2525
(C) 2424 (D) 2010
Answers
option B
Step-by-step explanation:
Given :-
x^2 -5x +1 = 0
To find:-
Find the value of (x^10 +1)/x^5 ?
Solution :-
Given quardratic polynomial is x^2 -5x +1 = 0
=> on dividing by x then
=> (x^2)/x -(5x/x) +(1/x) = 0/x
=> x-5+(1/x) = 0
=>x +(1/x) = 0+5
=> x + (1/x) = 5 ------------(1)
On squaring both sides then
=>[x+(1/x)]^2 = 5^2
=> x^2+(1/x)^2+2(x)(1/x) = 25
Since (a+b)^2 = a^2+2ab+b^2
=> x^2 + (1/x^2) +2 = 25
=>x^2+(1/x^2) = 25-2
=>x^2+(1/x^2) = 23 ----------(2)
On cubing (1) then
[x+(1/x)]^3 = 5^3
=> x^3+(1/x)^3+3(x)(1/x)[x+(1/x)] = 125
Since (a+b)^3 = a^3+b^3+3ab(a+b)
=> x^3+(1/x^3)+3[x+(1/x)] = 125
=> x^3+(1/x^3) +3(5) = 125
(from (1))
=> x^3+(1/x^3) +15 = 125
=> x^3+(1/x^3) = 125-15
=> x^3+(1/x^3) = 110--------(3)
On multiplying (2)&(3) then
[x^2+(1/x^2)][x^3+(1/x^3)] = 23×110
=> x^2[x^3+(1/x^3)]+(1/x^2)[x^3+(1/x^3)] = 2530
=> x^5+(x^2/x^3) +(x^3/x^2)+(1/x^5) = 2530
Since a^m ×a^n = a^(m+n)
=> x^5+(1/x^5)+x+(1/x) = 2530
=> x^5+(1/x^5)+5 = 2530
=> x^5+(1/x^5) = 2530-5
=> x^5+(1/x^5) = 2525
=> [x^5×x^5+1]/x^5 = 2525
=> (x^10+1)/x^5 = 2525
Answer:-
The value of (x^10+1)/x^5 for the given problem is 2525
Used formulae:-
- (a+b)^2 = a^2+2ab+b^2
- (a+b)^3 = a^3+b^3+3ab(a+b)
- a^m ×a^n = a^(m+n)
Answer:
option b
Step-by-step explanation:
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