Solving A Special Quartic | x^4 - 12x =17

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[bobbyheffley]

Ferrari's method - method 1
Descartes' method - method 2

[NadiehFan]

Unlike what you seem to think the nested square roots which appear in the solutions of this equation cannot be denested. For example, √(12√2 − 2) cannot be written as √p − √q for any rational p and q because the square of this is (p + q) − 2√(pq). We could take out a factor √2 to get ⁴√2·√(12 − √2) but √(12 − √2) cannot be denested either because 12² − (√2)² = 142 is not the square of a rational number. Also, it is not possible to write √(12√2 − 2) as ⁴√p − ⁴√q for any rational p and q because the square of this is √p + √q − 2·⁴√(pq) meaning that we would need to have pq = 1 and therefore q = 1/p, but there is no rational p such that √p + √(1/p) = 12√2.

Denesting nested square roots used to be a topic in high school algebra and perhaps it still is in countries like India, but in most countries those days seem to be long gone. A century ago any good high school student could denest a denestable square root easily using efficient techniques which were taught _and proved_ in class. But this is no longer the case and as a result such elementary problems now turn up in math Olympiads, which is quite ridiculous. But let's go back in time and discuss and prove a few theorems which were taught and proved in algebra classes.

*Theorem*

If a, b, c, d are positive rational numbers and √b and √d are irrational then a ± √b = c ± √d implies a = c and b = d.

Proof

a ± √b = c ± √d implies ±√b = (c − a) ± √d. Squaring both sides gives b = (c − a)² + d ± 2(c − a)√d and subtracting (c − a)² + d from both sides then gives (b − d) − (c − a)² = ±2(c − a)√d. Now the left hand side is rational, therefore the right hand side must also be rational. But √d is irrational, therefore the right hand side can only be rational if c − a = 0 which implies a = c. With c − a = 0 the equation (b − d) − (c − a)² = ±2(c − a)√d becomes (b − d) − 0 = 0 which implies b = d and this completes the proof.

Next, we have the following

*Theorem*

If a, b, c are positive rational numbers and √c is irrational, then there exist positive rational numbers p and q such that √(a ± b√c) = √p ± √q _if and only if_ a² − b²c is the square of a rational number. If d is a positive rational number such that a² − b²c = d² then p = (a + d)/2 and q = (a − d)/2.

Proof

Let a, b, c be positive rational numbers, √c irrational, and suppose there exist positive rational numbers p and q such that √(a ± b√c) = √p ± √q. Squaring both sides then gives a ± b√c = p + q ± 2·√p·√q or a ± √(b²c) = p + q ± √(4pq). Since √c and therefore √(b²c) is irrational the left hand side is irrational, therefore the right hand side must also be irrational. Since p and q are rational, p + q is rational, consequently √(4pq) must be irrational. In accordance with the previous theorem this implies a = p + q and b²c = 4pq. Consequently, a² − b²c = (p + q)² − 4pq = (p − q)² is the square of a rational number.

Conversely, if a, b, c are positive rational numbers, √c irrational, and a² − b²c is the square of a positive rational number d, then p = ½(a + d) and q = ½(a − d) are positive rational numbers with p > q and such that p + q = a and 4pq = b²c so 2·√p·√q = b√c which implies a ± b√c = p + q ± 2·√p·√q = (√p ± √q)² so √(a ± b√c) = √p ± √q which completes the proof.

In short, if if a, b, c are positive rational numbers, √c irrational, and d is a positive rational number such that d² = a² − b²c then

√(a ± b√c) = √((a + d)/2) ± √((a − d)/2) = ½√(2a + 2d) ± ½√(2a − 2d)

[scottleung]

Got 'em all (although I went a little further than you did in terms of denesting the solutions)!

[shamicray]

but sir i think the answer is not correct as if you put the values of x that you found and put it in the equation there is coming an error of so i think there is no real solution of the sum.