Question

why we took aquare of \(x-\dfrac1{x}\) in question ...(1) and don't take square in question ...(2) as shown in below attachement

Answer 1

@kc_kennylau @kropot72 @kohai @kirbykirby halp please..

Answer 2

@bibby @skullpatrol @shamil98 @satellite73 @adrynicoleb @Ashleyisakitty @Agentjamesbond007 @alyssa_michelle1996

Answer 3

I really don't know.
does this have anything to do with absolute values

Answer 4

It seems to me that you do square t in the second one, even though it is written as t. There are quite a lot of typos in the document....

Answer 5

i don't know @triciaal

Answer 6

I am thinking it has something to do with the positive or negative value of x
question 1 when x is +ve possible 2 +ve and 2-ve roots
when x is -ve possible 2 +ve and 2 -ve roots
question 2 when x is +ve possible 4 +ve roots when x is -ve possible 4 -ve roots

Answer 7

@skullpatrol

Answer 8

@jim_thompson5910

Answer 9

@wolf1728

Answer 10

@tHe_FiZiCx99 @thomaster

Answer 11

@Compassionate

Answer 12

@anjum @asnaseer

Answer 13

Looking at the problem again, I think it is asking why don't we substitute the whole t^2 expression into the formula in problem 2 as we do in problem 1. Do you think I am understanding the question correctly?

Answer 14

@nincompoop @Nikato, @aum, @Zepdrix , @satellite73, @mathstudent55 can any of you help with this?

Answer 15

To reduce confusion, t should be equal to \[x-\frac{ 1 }{ x }\] in both problems. At least the way I see it, it is a typo.

Answer 16

And \[t ^{2}=x ^{2}+\frac{ 1 }{ x ^{2} }-2\] for both of them too.

Answer 17

I think it has to do with eliminating any complex roots so we have -1 and not working with rt of -1 somehow

Answer 18

@triciaal @JoannaBlackwelder when we see question 1 here
we see they take square of \(x-\dfrac1{x}\)=\(x^2+\dfrac1{x^2}-2\)
but when we see in question 2
we see that here if we take square of \(x-\dfrac1{x}\)=it should be \(x^2+\dfrac1{x^2}-2\) but instead of it they are taking square of \(x^2+\dfrac1{x^2}+2=t^2\)

Answer 19

It looks like another typo to me.

Answer 20

typo mistake is not @joannaBlackwelder

Answer 21

It leaves off the square on the t^2 in problem 2 too. It looks like a bunch of typos. Which makes it very difficult to follow.

Answer 22

It seems like it comes down to making the equation easier to simplify.

Answer 23

Yah looks like a bunch of typos D:
Like in problem 1, the substitution is supposed to be \(\Large\rm t=x-\frac{1}{x}\)

Answer 24

note that for question 1 t = x + 1/x but for question 2 t = x - 1/x

question 1 we squared to get a quadratic but for question 2 when we get rid of fraction we already have a quadratic x^2 -1 ( difference of 2 squares)

Answer 25

the other way around: for question 1, t = x - 1/x but for question 2, t = x + 1/x

Answer 26

yes...@triciaal i think they changes in question 1 x-1/x into square which becomes \(x^2 +\dfrac1{x^2}-2\)
but in question 2
they didin't changes x-1/x into square ..this is the thing i want to get clear myself

Answer 27

if we take square in question 2 then 2 would be negative but here is 2 positive in question 2

Answer 28

They did square t in question 2. They just left off the ^2 and they had a typo.

Answer 29

\(2x^4 + 3x^3 - 4x^2 - 3x + 2 = 0\)
Notice the symmetry in the absolute values of the coefficients.
Divide throughout by \(x^2\)
\(2x^2 + 3x - 4 - 3/x + 2/x^2 = 0\)
\(2(x^2 + 1/x^2) + 3(x - 1/x) - 4 = 0 \)
\(2(x^2 + 1/x^2 - 2 + 2) + 3(x - 1/x) - 4 = 0 \)
\(2\{(x-1/x)^2 + 2\} + 3(x - 1/x) - 4 = 0 \)
\(2(x-1/x)^2 + 4 + 3(x - 1/x) - 4 = 0 \)
Let \(t = x - 1/x\)
\(2t^2 + 3t = 0\)

Answer 30

\(6x^4-35x^3+62x^2-35x+6 = 0\)

Divide throughout by \(x^2\)
\(6x^2 - 35x + 62 - 35/x + 6/x^2 = 0\)
\(6(x^2 + 1/x^2) - 35(x+1/x) + 62 = 0\)
Notice the second term here has \(x + 1/x\) but the previous one had \((x-1/x)\)

Answer 31

\(6(x^2 + 1/x^2 + 2 - 2) - 35(x + 1/x) + 62 = 0\)
\(6\{(x+1/x)^2 - 2\} - 35(x + 1/x) + 62 = 0\)
\(6(x+1/x)^2 - 12 - 35(x + 1/x) + 62 = 0\)
Let \(t = (x + 1/x)\)
\(6t^2 - 12 - 35t + 62 = 0\)
\(6t^2 - 35t + 50 = 0\)

Answer 32

@aum thanks

Answer 33

You are welcome.

Answer 34

@aum i don't understand how u do..?

Answer 35

@sammixboo