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

[@bobbyheffley4955]

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)

[@scottleung9587]

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 0.0000011...... so i think there is no real solution of the sum.

[@NadiehFan]

Calculators always have finite precision and therefore rounding errors. You should _never_ use a calculator to verify a solution of an equation. Always use pen and paper and algebra to verify solutions.

[@shamicray]

yes but sir if you calculate using a calculator it should give errors like 0.0000000......1 then they get limited out so there was an error . moreover sir i can explain it using some algebra that the solution had minute errors. @@NadiehFan

[@NadiehFan]

@@shamicray Can you give the exact solution you tested? Let's take the positive real solution found in the video at 6:00 which is x = (√2 + √(12√2 − 2))/2 Squaring this we have x² = (2 + 12√2 − 2 + 2·√2·√(12√2 − 2))/4 = (12√2 + 2√(24√2 − 4))/4 and again squaring this we have x⁴ = (288 + 4((24√2 − 4) + 2·12√2·2√(24√2 − 4))/16 = (288 + 16(6√2 − 1) + 48√(48√2 − 8))/16 = 18 + 6√2 − 1 + 3√(48√2 − 8) = 17 + 6√2 + 6√(12√2 − 2) and we also have 12x = 12((√2 + √(12√2 − 2))/2) = 6√2 + 6√(12√2 − 2) Consequently, we find x⁴ − 12x = (17 + 6√2 + 6√(12√2 − 2)) − (6√2 + 6√(12√2 − 2)) = 17 So, x = (√2 + √(12√2 − 2))/2 is indeed an exact root of the equation x⁴ − 12x = 17 You can do the same verification for the negative root x = (√2 − √(12√2 − 2))/2 which will then give x⁴ = 17 + 6√2 − 6√(12√2 − 2) 12x = 6√2 − 6√(12√2 − 2) so we again find that x⁴ − 12x = (17 + 6√2 − 6√(12√2 − 2)) − (6√2 − 6√(12√2 − 2)) = 17 which verifies that x = (√2 − √(12√2 − 2))/2 is also an exact root of the equation x⁴ − 12x = 17 I conclude that your claim that the roots found by SyberMath are not exact is false.

[@shamicray]

there is no real solution to the problem @@NadiehFan

[@NadiehFan]

@@shamicray You are wrong. The equation has two real solutions and I proved to you that the real solutions found in the video are correct.