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Questions 3 and 6 plz help tuff questions of numbers system

Questions 3 and 6 plz help tuff questions of numbers system

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3 Answers

Vedant
35 Points
2 years ago
Dear student,
x^3 + 1/x^3 – 4(x^2 + 1/x^2) + x + 1/x can also be written as
(x + 1/x)^3 – 2(x + 1/x) – 4(x^2 + 1/x^2)
Vedant
35 Points
2 years ago
I am sorry
It was accidentally posted    
Here is the complete version:
Dear student,
x^3 + 1/x^3 – 4(x^2 + 1/x^2) + x + 1/x can also be written as
(x + 1/x)^3 – 2(x + 1/x) – 4(x^2 + 1/x^2) (recall the binomial expansion of (x + 1/x)^3)
x^2 + 1/x^2 = (x + 1/x)^2 – 2
This implies that    – 4(x^2 + 1/x^2) = – 4((x + 1/x)^2 – 2) = – 4((x + 1/x)^2) + 8
Therefore the whole equation can be written as:
(x + 1/x)^3 – 2(x + 1/x) – 4((x + 1/x)^2) + 8
Now substituting x + 1/x as p, we have
p^3 – 2p – 4p^2 + 8
= p^3 – 4p^2 – 2p + 8
= p^2(p – 4) – 2(p – 4)
 = (p – 4) (p^2 – 2)
Now we have to calculate the value of p
We have been given that x = sqrt(7 – 4sqrt(3)) = sqrt(2^2 + sqrt(3)^2 – 2 * 2sqrt(3))
= sqrt([2 – sqrt(3)]^2)
= 2 – sqrt(3)
x = 2 – sqrt(3)
1/x = 1/{2 – sqrt(3)}
On rationalising the denominator, we get
1/x = 2 + sqrt(3)
p = x + 1/x
= 2 – sqrt(3) + 2 + sqrt(3) = 4
On substituting p = 4 in (p – 4) (p^2 – 2)
We get (p – 4) (p^2 – 2) = 0
THAT’S THE ANSWER FOR THE THIRD QUES
ANSWER FOR SIXTH QUES IS COMING SOON
 
Regards
Vedant
 
Vedant
35 Points
2 years ago
HERE’S ANS 6 :
We want the minimum value of a + b + c and we know that a,c is not equal to 0
We try putting b = 0
the equation becomes (a0)^2 =  cc0
this leads to a contradiction as (a0)^2 = (10a)^2 = 100a^2 = (a^2)00 which does not have one zero at the end as in cc0
Therefore we try putting b= 1 in the equation
this becomes (a1)^2 = cc1
As we can’t have a = 0, we put a = 1
we get 11^2 = cc1
we know that 11^2 = 121 and 1 and 2 are not the same numbers
Let’s try the next case
a = 2, we get
21^2 = cc1 which is indeed true because 21^2 = 441 and 4=4
Therefore we get our answer as 21^2 = 441
which implies that a = 2, b = 1 and c = 4
a + b + c
= 2 + 1 + 4
= 7 Ans

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