how do you solve this quartic?

TeeEm

Any thoughts?

A stab in the dark but the fact that the roots (for the same equation but equal to 0) are all 4 apart, might be useful somehow? Not sure how though ahaha

(edited 8 years ago)

Reply 2

TeeEm

OP

19

Original post by IIEquinoxII

A stab in the dark but the fact that the roots (for the same equation but equal to 0) are all 4 apart, might be useful somehow? Not sure how though ahaha

might be important... I did not spot that

Reply 3

Andy98

20

Original post by TeeEm

Any thoughts?

Would it not be the normal way?

Reply 4

IIEquinoxII

9

Original post by TeeEm

might be important... I did not spot that

Well since, it intersects with y=1680, you can assume the smallest solution is less than -5, the middle two are between -1 and 3 and the last one is greater than 7.

And since there is symmetry, we can assume they have all shifted from -5,-1,3 and 7 by the same distance?

^Scratch that, if the middle two solutions are beyond the point of where dy/dx goes from >1 to <1 the amount shifted will be different

(edited 8 years ago)

Reply 5

TeeEm

OP

19

Original post by Andy98

Would it not be the normal way?

Normal way is good but I am looking for some clever "trick"
what if it does not factorize?

Reply 6

Asuna Yuuki

9

Original post by TeeEm

Any thoughts?

Well, multiplying it out gives you

x⁴- 4x³ - 34x² + 76x - 1575 = 0

For experimentation's sake, I'll try taking out a factor of (x - 7)

... and that gets me nowhere.

Honestly, I was just clutching at straws. Who's next?

Reply 7

Andy98

20

Original post by TeeEm

Normal way is good but I am looking for some clever "trick"
what if it does not factorize?

According to Wiki there's a formula

Spoiler

Reply 8

TeeEm

OP

19

Original post by Andy98

According to Wiki there's a formula

Spoiler

I am aware of that but this is non calculator

Reply 9

RMNDK

3

Original post by TeeEm

I am aware of that but this is non calculator

Ok this is a really crazy and possibly stupid way to do this...

But, do prime factor decomposition of 1680.

If you notice, your factors are (x+a), where a is an odd number. This means value of the brackets must be even. Using your factors from factor decomposition, form even factors. You can then cross-reference this to your brackets and figure out what value of x it is.

(Sounds horrible)

Reply 10

Andy98

20

Original post by TeeEm

I am aware of that but this is non calculator

Ahhh that's a shame.

Hmmmm....Using the expanded out form kindly given by Asuna

Original post by Asuna Yuuki

x⁴- 4x³ - 34x² + 76x - 1575 = 0

There's the possibility of factor theorem, although it looks very time consuming. Realistically it'd be nice numbers in a non-calculator question.

Original post by TeeEm

Although, looking at the original question; we could be dramatically overthinking and it could be as simple as: adding the 7 and the 3, and subtracting the 5 and the 1 for the respective roots.

Reply 11

TeeEm

OP

19

Original post by RMNDK

Ok this is a really crazy and possibly stupid way to do this...

But, do prime factor decomposition of 1680.

If you notice, your factors are (x+a), where a is an odd number. This means value of the brackets must be even. Using your factors from factor decomposition, form even factors. You can then cross-reference this to your brackets and figure out what value of x it is.

(Sounds horrible)

this is a good idea and it would work well if for instance if this quartic came from a problem that had lets say integer solutions

Reply 12

RMNDK

3

Original post by TeeEm

this is a good idea and it would work well if for instance if this quartic came from a problem that had lets say integer solutions

What do you mean? It does have an integer solution?

Reply 13

TeeEm

OP

19

Original post by Andy98

Ahhh that's a shame.

Hmmmm....Using the expanded out form kindly given by Asuna

There's the possibility of factor theorem, although it looks very time consuming. Realistically it'd be nice numbers in a non-calculator question.

Although, looking at the original question; we could be dramatically overthinking and it could be as simple as: adding the 7 and the 3, and subtracting the 5 and the 1 for the respective roots.

i think the gap between the number might not be coincidental.

Reply 14

Andy98

20

Original post by TeeEm

i think the gap between the number might not be coincidental.

Yeah, there must be something

Reply 15

Asuna Yuuki

9

Original post by TeeEm

I am aware of that but this is non calculator

I'm unaware of any method better than using trial and error to find a root of the equation

x⁴- 4x³ - 34x² + 76x - 1575 = 0

I tried substituting in 9, which gives 6561 - 4*729 - 34*81 + 76*9 - 1575 = 0

Therefore (x - 9) is a factor of x⁴- 4x³ - 34x² + 76x - 1575 = 0

Dividing gives (x - 9)(x³ + 5x² + 11x + 175) = 0

After trying a few more values, x³ + 5x² + 11x + 175 = 0 when x = -7

Dividing again gives (x - 9)(x + 7)(x² - 2x + 25) = 0

The quadratic x² - 2x + 25 = 0 has no real roots.

Therefore, your solutions are x = -7, x = 9.

Sorry for posting the whole thing. I got carried away...

Reply 16

username1560589

20

Original post by TeeEm

Any thoughts?

I did it by saying b = x - 1, then using quadratic formula on the quadratic in b^2. Gives you answers but not looking nice to simplify so far.

Reply 17

TeeEm

OP

19

Original post by Asuna Yuuki

I'm unaware of any method better than using trial and error to find a root of the equation

x⁴- 4x³ - 34x² + 76x - 1575 = 0

I tried substituting in 9, which gives 6561 - 4*729 - 34*81 + 76*9 - 1575 = 0

Therefore (x - 9) is a factor of x⁴- 4x³ - 34x² + 76x - 1575 = 0

Dividing gives (x - 9)(x³ + 5x² + 11x + 175) = 0

After trying a few more values, x³ + 5x² + 11x + 175 = 0 when x = -7

Dividing again gives (x - 9)(x + 7)(x² - 2x + 25) = 0

The quadratic x² - 2x + 25 = 0 has no real roots.

Therefore, your solutions are x = -7, x = 9.

Sorry for posting the whole thing. I got carried away...