Question

help please ! any idea ???

Answer 1

solve this system

\[[x^2 +(x^2 /2y)^2 ]/2y = y-\sqrt{P}\]

y^2 = P +x^2

my work :

x^2 = y^2 -P
x^2 =(y-sqrt(p))(y+sqrt(p))

{(y-sqrt(p))(y+sqrt(p)) +{[(y-sqrt(p)(y+sqrt(p))]/2y}^2 }/2y = (y-sqrt(p))

(y-sqrt(p))(y+sqrt(p)) +[(y-sqrt(p))(y+sqrt(p))/2y]^2 = 2y(y-sqrt(p))

(y-sqrt(p))(y+sqrt(p)) +[(y-sqrt(p))^2(y+sqrt(p))^2]/4y^2 = 2y(y-sqrt(p))

4y^2(y-sqrt(p))(y+sqrt(p)) +(y-sqrt(p)^2(y+sqrt(p))^2 = 8y^3(y-sqrt(p)) / (y-sqrt(p))

4y^2(y+sqrt(p)) +(y-sqrt(p))(y+sqrt(p))^2 = 8y^3

4y^3 +4sqrt(p)y^2 +(y-sqrt(p))(y^2 +2sqrt(p)*y +p) = 8y^3

4y^3 +4sqrt(p)y^2 +y^3 +2sqrt(p)y^2 +py - sqrt(p)y^2 -2py -psqrt(p) = 8y^3

5y^3 +5sqrt(p)y^2 -py -psqrt(p) = 8y^3

3y^3 -5sqrt(p)y^2 +py +psqrt(p) =0

any idea to continue it from this ???

@Hero
please !

Answer 2

@Vocaloid

Answer 3

@Angle whenever you get a chance, this is a little above my skill level

Answer 4

@Ultrilliam please

Answer 5

@Hero If you can, this is above my teachable skill level

Answer 6

@AngeI
@Nnesha

please any idea ?

Answer 7

Solution Attempt: (Assuming p is a natural number helps with organizing the terms)

Given:
\((a)\) \(\dfrac{x^2 + \left(\dfrac{x^2}{2y}\right)}{2y} = y - \sqrt{p}\)

\((b)\) \(p = y^2 - x^2\)

The solution steps are as follows:
1. Expand \(\left(\dfrac{x^2}{y^2}\right)^2\) to get:
\(\dfrac{x^2 + \dfrac{x^4}{4y^2}}{2y} = y - \sqrt{p}\)

2. Multiply top and bottom of big fraction by \(4y^2\):
\(\left(\dfrac{x^2 + \dfrac{x^4}{4y^2}}{2y}\right) \left(\huge\dfrac{4y^2}{4y^2}\right)= y - \sqrt{p}\)

3. Distribute:
\(\dfrac{4y^2\left(x^2 + \dfrac{x^4}{4y^2}\right)}{4y^2(2y)} = y - \sqrt{p}\)


\(\dfrac{4x^2y^2 + x^4}{8y^3}=y - \sqrt{p}\)

4. Multiply both sides by \(8y^3\):
\(4x^2y^2 + x^4 = 8y^4 - (8\sqrt{p})y^3\)

5. Factor out \(x^2\) on the LHS:
\(x^2(4y^2 + x^2)=8y^4 - (8\sqrt{p})y^3\)

6. This would be a good step to replace \(x^2\) with \(y^2 - p\) from equation \((b)\):
\((y^2 - p)(4y^2 + y^2 - p) = 8y^4 - (8\sqrt{p})y^3\)

\((y^2 - p)(5y^2 - p) = 8y^4 - (8\sqrt{p})y^3\)

7. FOIL (I hate FOIL but would rather FOIL than do distributive property in this instance:
\(5y^4 - (6p)y^2 + p^2 = 8y^4 - (8\sqrt{p})y^3\)

8. Combine like terms to get:
\(3y^4 - (8\sqrt{p})y^3 + (6p)y^2 - p^2 = 0\)

The extra \(-p^2\) is what makes it inconvenient to continue solving this.

Answer 8

If you can think of a way to get rid of that \(-p^2\) I can finish solving it @jhonyy9

Answer 9

Actually, there may be a way. Completing the square might work. I'll try that next.

Answer 10

completing the square using what terms ?

Answer 11

do you think getting a square perfect using every terms ?

Answer 12

Yeah, I don't know if it will really work.

Answer 13

The first three terms.

Answer 14

If we can express the first three terms in some form of a square binomial then we will finally be able to solve it completely

Answer 15

That is assuming all my steps up to this point are correct.

Answer 16

i think that are correct bc. i ve cheked step by step

Answer 17

Because in actuality, we can factor \(y^2\) from the first three terms to get:

Answer 18

\(3y^4 - (8\sqrt{p})y^3 + (6p)y^2 - p^2 = 0\)
\(y^2(3y^2 - (8\sqrt{p})y + 6p) - p^2 = 0\)

Answer 19

Notice that now \(3y^2 - (8\sqrt{p})y + 6p\) is a quadratic trinomial that we can complete the square with.

Answer 20

Our goal is to eventually get some form of \(a^2b^2 - p^2 = 0\) which we can then express as \((ab)^2 - p^2 = 0\) Then then use difference of squares to express the result as \((ab + p)(ab - p) = 0\) Then use the zero product property afterwards:
\(ab + p = 0\) and \(ab - p = 0\)

Answer 21

first term is ((srt3)y)^2 and the last one is (sqrt(6p))^2

Answer 22

You're losing me there bro. That's not the approach I'm taking with this. I don't immediately resort to roots if I can avoid them.

Answer 23

i understand you but i ve thought a square perfect in this form (a-b)^2

Answer 24

Yes it is. It can be some form of that. But again, you don't need to resort to rooting the coefficients to get there. All you need to do is complete the square on the quadratic trinomial.

Answer 25

When you complete the square you add \(\left(\dfrac{b}{2}\right)^2\) to both sides

Answer 26

I realize this may not work because we may end up with an extra term we don't need.

Answer 27

Yeah, nevermind that idea.

Answer 28

this -(8sqrtp) can we getting in this necessary form of -2ab ?

so -2*4sqrtp when we know that the first term is ((sqrt3)y)^2 and the last one is ((sqrt6)p)^2

Answer 29

using formule (a-b)^2 = a^2 -2ab +b^2

Answer 30

i think dont function bc. srt3*srt6 =sqrt18 =sqrt(9*2) = 3sqrt2

Answer 31

now using this terms we get -3sqrt2p when we need getting -8sqrtp

Answer 32

I don't like this use of roots. I don't solve using those methods.

Answer 33

If you want to solve it that way, be my guest.

Answer 34

ok - but than what my be your idea that can using here ?

Answer 35

I'll have to get back to you on that.

Answer 36

I'll have to toy and play around with it a bit more. But if there really is a way to solve it, I'll figure it out.

Answer 37

ok thank you but i will try again than i get any idea - but i wait your way too - so than not in this night (here) sorry - there is afternoon - yes ? so than tomorow

Answer 38

I'll get back to you as soon as I can. Maybe tomorrow who knows.

Answer 39

ok - great - thank you -

Answer 40

@Hero than we consider this 3y^2 -(8sqrtp)y +6p = 0 a quadratic in yso we get delta -the discriminant smaler than zero ,i ve got discriminant equal -8p so in this case this mean that has complex roots - yes ? than how can we factoriz it in a square perfect ?

Answer 41

bc we know again that p is a natural number greater than zero