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Equations Reducible to Quadratics (1 Viewer)

random-1006

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ummm dude...i'm getting m^2 + 4 - 0 which is incorrect as there is no b. o_O?

i did (x)^e + 1/ x^e + 3 = 0

X everything by x^e

(x^e)^2 + 1 + 3 = 0

let m = x^e

m^2 + 4 = 0 ...?
thats wrong

u= e^x
u + 1/u + 3 =0
u^2 +1 +3u=0
u^2+3u +1 =0
u= -3 +- sqrt( 9-4) / 2
u=( -3 + sqrt5 )/ 2 or u = (-3 - sqrt5) / 2

but u= e^x ---> x=lnu

so take ln of the two things above , but they are both negative, you cannot take log of negative

so there are no solns for this eqn
 
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random-1006

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I remember seeing that question in a 2 unit paper but it was logx instead of e^x

all these equations reducible to quadratics are of the same form, just different variable, they can have e^x, lnx , sin, cos or tan, they are fairly easy to do onc you have seen a few
 

MrBrightside

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all these equations reducible to quadratics are of the same form, just different variable, they can have e^x, lnx , sin, cos or tan, they are fairly easy to do onc you have seen a few
ahhh shit i was reading the e^x the wrong way around...i thought the "e" went up ffs.

anyways thanks really appreciate your help...i understand this topic clearer now :)

here is one for the road --- I did it but stuffed up somewhere -_-

Solve:

(x^2 + 1 / x^2) - 9(x^2 + 1 / x^2) + 20 = 0

correct to 2 decimal places.
(x cannot = 0).

PS: is it easier to use that maths writing thing for these equations on this website? Umm and can i have ur msn plz :) for when BOSF goes down :(? and which textbook do you recommend for 2U maths? and do you think i shud get a diff one for yr 12?....because maths in focus is not the best to understand things.
 

random-1006

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ahhh shit i was reading the e^x the wrong way around...i thought the "e" went up ffs.

anyways thanks really appreciate your help...i understand this topic clearer now :)

here is one for the road --- I did it but stuffed up somewhere -_-

Solve:

(x^2 + 1 / x^2)^2 - 9(x^2 + 1 / x^2) + 20 = 0 i added in the square on first term, think you made mistake.

correct to 2 decimal places.
(x cannot = 0).

PS: is it easier to use that maths writing thing for these equations on this website? Umm and can i have ur msn plz :) for when BOSF goes down :(? and which textbook do you recommend for 2U maths? and do you think i shud get a diff one for yr 12?....because maths in focus is not the best to understand things.
u= x^2 + 1/x^2 , u^2 -9u +20 = 0
(u-5)(u-4)=0
u= 5, u= 4

x^2 + 1/x^2 = 5
x^4 + 1 = 5x^2
x^4 -5x^2 +1=0 ( let m ( different letter) = x^2 )
m^2 -5m +1=0
m= [5 +- sqrt( 21) ] / 2 = x^2 ( then take sqrt of both sides )

m= 4.8 or m= 0.208

then x^2 = 4.8 and x^2 = 0.208

remember when you take sqrt you get +- , so this first eqn will give 4 solns

two solutions from this one

x^2 +1/x^2 =4
x^4 +1 = 4x^2 ( let n= x^2)
n^2-4n +1= 0
n= ( 4 +-sqrt(16-4) ) / 2 = x^2, repeat same as above

Im tired, im going to sleep lol, i dnt have uni tommorow lol
 
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MrBrightside

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u= x^2 + 1/x^2 , u^2 -9u +20 = 0
(u-5)(u-4)=0
u= 5, u= 4

x^2 + 1/x^2 = 5
x^4 + 1 = 5x^2
x^4 -5x^2 +1=0 ( let m ( different letter) = x^2 )
m^2 -5m +1=0
m= [5 +- sqrt( 21) ] / 2 = x^2 ( then take sqrt of both sides )

m= 4.8 or m= 0.208

then x^2 = 4.8 and x^2 = 0.208

remember when you take sqrt you get +- , so this first eqn will give 4 solns

two solutions from this one

x^2 +1/x^2 =4
x^4 +1 = 4x^2 ( let n= x^2)
n^2-4n +1= 0
n= ( 4 +-sqrt(16-4) ) / 2 = x^2, repeat same as above
ahh yes its squared sry lol, i think my minds going...anyways thx, i gtg get some sleep before i drop...Night man. :)
 

hscishard

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PS: is it easier to use that maths writing thing for these equations on this website? Umm and can i have ur msn plz :) for when BOSF goes down :(? and which textbook do you recommend for 2U maths? and do you think i shud get a diff one for yr 12?....because maths in focus is not the best to understand things.
No offence..but maths in focus is brief and concise.
 

random-1006

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No offence..but maths in focus is brief and concise.

just get a book of past papers or something to look at and go through, and when you find a question you dont understand ask your maths teacher at the end of your next lesson, or go and annoy them at lunch, look at your school library, at my library there was a fair few hsc maths study guides/ exam paper books that hardly anyone else used that i could borrow.
 

MrBrightside

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u= x^2 + 1/x^2 , u^2 -9u +20 = 0
(u-5)(u-4)=0
u= 5, u= 4

x^2 + 1/x^2 = 5
x^4 + 1 = 5x^2
x^4 -5x^2 +1=0 ( let m ( different letter) = x^2 )
m^2 -5m +1=0
m= [5 +- sqrt( 21) ] / 2 = x^2 ( then take sqrt of both sides )

m= 4.8 or m= 0.208

then x^2 = 4.8 and x^2 = 0.208

remember when you take sqrt you get +- , so this first eqn will give 4 solns

two solutions from this one

x^2 +1/x^2 =4
x^4 +1 = 4x^2 ( let n= x^2)
n^2-4n +1= 0
n= ( 4 +-sqrt(16-4) ) / 2 = x^2, repeat same as above

Im tired, im going to sleep lol, i dnt have uni tommorow lol
umm yea bro...i did the exact same thing as u, got the exact same answers but the book has totally diff ans...

book saids x = -+2.19, -+0.46, -+1.93, -+0.52

i got this: x = -+4.79, -+0.21, -+3.73, -+0.27 (rounded to 2 dec place)


so yea i dunno if u did something wrong there?:confused:
 

random-1006

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umm yea bro...i did the exact same thing as u, got the exact same answers but the book has totally diff ans...

book saids x = -+2.19, -+0.46, -+1.93, -+0.52

i got this: x = -+4.79, -+0.21, -+3.73, -+0.27 (rounded to 2 dec place)


so yea i dunno if u did something wrong there?:confused:

these highlighted answers from the book are from the two eqns i shown above

it is most likely calculator work , make sure you use brackets when you enter large expressions with fractions into your calculator

EDIT lol, you havent taken sqrt of them!!, take sqrt of all your answers, and you get the answers

you have figured out x^2 to be 4.79, 0.21, 3.73 and 0.27 ( note the +- comes when you take the sqrt)
 
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MrBrightside

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these highlighted answers from the book are from the two eqns i shown above

it is most likely calculator work , make sure you use brackets when you enter large expressions with fractions into your calculator

EDIT lol, you havent taken sqrt of them!!, take sqrt of all your answers, and you get the answers

you have figured out x^2 to be 4.79, 0.21, 3.73 and 0.27 ( note the +- comes when you take the sqrt)
ooooooooooooooo thx man ur the best :)...thank god i did everything else right :)...just forgot the last bit..thx again! :D:sun::sun::sun::haha:
 

MrBrightside

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umm hey i have another qs :(...

Solve for 0 <= x <= 360

3tan^4x - 10tan^2 + 3 = 0

now this is what i did.

i let m= tan^2x and i went from there to get only two ans

30, and 60....these r correct although there are like 6 other answers that i have no idea how they got...did i miss something..again?
 

random-1006

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umm hey i have another qs :(...

Solve for 0 <= x <= 360

3tan^4x - 10tan^2 + 3 = 0

now this is what i did.

i let m= tan^2x and i went from there to get only two ans

30, and 60....these r correct although there are like 6 other answers that i have no idea how they got...did i miss something..again?

when you take sqrt you get +- , so you will go to all four quadrants
u= tan^2

3u^2 -10u +3=0
u= 10 +- sqrt (100- 4(3)(3)) / 2

we end up taking sqrt of negative, you have written eqn down wrong
 
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hscishard

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umm hey i have another qs :(...

Solve for 0 <= x <= 360

3tan^4x - 10tan^2 + 3 = 0

now this is what i did.

i let m= tan^2x and i went from there to get only two ans

30, and 60....these r correct although there are like 6 other answers that i have no idea how they got...did i miss something..again?
When you square root the tan^2x, you must put +/- on the RHS. There should be 8 answers.
 

deterministic

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when you take sqrt you get +- , so you will go to all four quadrants
u= tan^2

3u^2 -10u +3=0
u= 10 +- sqrt (100- 4(3)(3)) / 2

we end up taking sqrt of negative, you have written eqn down wrong
Lol... u= (10 +- sqrt (100- 4(3)(3))) / 6 not u= (10 +- sqrt (100- 4(3)(3))) / 2

OR factorise
(3u-1)(u-3)=0
and yes there should be 8 solutions
 

MrBrightside

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yea i got u= 3 or 1/3

tan^2x = sqrt3 or tan^2x = 1 / 3

shift tan sqrt3 or shift tan 1/3

x = 60 x = 30

where do I go from there?
 

deterministic

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(tanx)^2=1/3 or 3
so tanx=1/sqrt(3) or -1/sqrt(3) or sqrt(3) or -sqrt(3)
tanx=1/sqrt(3)

means x=30 or 210 (remember 0<x<360)
and do the same with the rest
 

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