# What are symmetric equations

Joachim Mohr Mathematics Music Delphi ### 4th degree equation. - Special cases Decomposition: x4+ ax3+ bx2+ cx + d = (x2+ px + q) (x2+ sx + t)

Previous: 4th degree equations, general.
Special cases of the equation of the 4th degree x4 + ax3 + bx2 + cx + d = 0 presented with a simple solution formula:

Special case I 4ab-a3= 8c (the diagram is symmetrical)

Special case II c = 0 and 4b = a2

Special case III p = s with further special conditions

Special case IV q = t

Special case V Symmetrical equation

Special case VI Biquadratic equation x4 + bx2 + d = 0 (a = 0 and c = 0)

### I special case 4ab-a3= 8c or 2p-a = 0

The diagram of the function x-> x4+ ax3+ bx2+ cx + d is symmetrical to the straight line with the equation x = - a / 4 parallel to the y-axis.
If you substitute x = u - a / 4, you get a biquadratic equation the shape u4+ b'u2+ d '= 0

#### Short version:

In this case the solution is ap = - 2 ———————— 1/2 2 q = - (c + \ / c - adaas = - 2 ———————— 1/2 2 t = - (c - \ / c - ada solution can be copied: p = 1/2 * aq = 1 / a * (c + sqrt (c ^ 2-a ^ 2 * d) or q = c / a + sqrt (c ^ 2 / a ^ 2-d) s = 1/2 * at = 1 / a * (c-sqrt (c ^ 2-a ^ 2 * d) or t = c / a-sqrt (c ^ 2 / a ^ 2-d) x1 = -p / 2 + sqrt ((p / 2) ^ 2-q) x2 = -p / 2-sqrt ((p / 2) ^ 2-q) x3 = -s / 2 + sqrt ((s / 2) ^ 2-t) x4 = -s / 2-sqrt ((s / 2) ^ 2-t) Note: 4ab-a3-8c is invariant to linear transformations, i.e.
from (u-t)4+ a1(u-t)3+ b1(u-t)2+ c1(u-t) + d = x4+ ax3+ bx2+ cx + d for x = u-t
follows 4a1b1 - a13 - 8c1 = 4ab - a3 - 8c

### Detailed:

Examples: x ^ 4 + 4x ^ 3 + 6x ^ 2 + 4x + 1 = 0 a = 4 b = 6 c = 4 Here: 4ab-a3= 8c (= 32) x ^ 4 + 2x ^ 3-11x ^ 2-12x + 36 = 0 a = 2 b = -11 c = -12 Here: 4ab-a3= 8c (= -96) In the calculation scheme of the general case, when calculating
q divided by (-a + 2p).
In the calculation it must not be there that a = 2p. In the case a = 2p then p = s = a / 2 and c = 1 / 2ab-1 / 8a3 or 4ab-a3= 8c. The calculation of p, q, r and s as well as the zeros is then short: Namely: p = 1/2 * a; s = 1/2 * a; q = (4 * b-a ^ 2 + sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; t = (4 * b-a ^ 2-sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; x1 = -p / 2 + sqrt ((p / 2) ^ 2-q); x2 = -p / 2-sqrt ((p / 2) ^ 2-q); x3 = -s / 2 + sqrt ((s / 2) ^ 2-t); x4 = -s / 2-sqrt ((s / 2) ^ 2-t); And: The diagram is then symmetrical to a parallel line of the y-axis.

Derivation: From x4+ a * x3+ b * x2+ cx + d = (x2+ px + q) (x2+ sx + t) follows:
a = p + s, b = q + ps + t, c = qs + pt, d = qt. With a = 2p and s = p it follows:
a = 2p, b = p ^ 2 + q + t, c = pq + pt, d = qt. Further transformations:
p = a / 2, b = 1 / 4a ^ 2 + q + t, c = 1 / 2aq + 1 / 2at.
Solve the second equation for t and insert it into the third:
t = b-1 / 4a ^ 2-q, c = 1 / 2aq + 1 / 2a (b-1 / 4a ^ 2-q) = 1 / 2ab-1 / 8a ^ 3 q.e.d

And: q is then the solution of the equation: 4q ^ 2 + (a ^ 2-4b) q + 4d = 0

Calculation with Maple f: = x-> x ^ 4 + a * x ^ 3 + b * x ^ 2 + (4 * a * b-a ^ 3) / 8 * x + d; g: = x-> x ^ 2 + p * x + q; s: = a-p; t: = b-q-a * p + p ^ 2; h: = x-> x ^ 2 + s * x + t; r: = x -> ((4 * a * ba ^ 3) / 8-b * p + a * p ^ 2-p ^ 3-a * q + 2 * p * q) * x + (db * q + q ^ 2 + a * p * qp ^ 2 * q); simplify (f (x) -g (x) * h (x) -r (x)); # result 0 solve ({(4 * a * ba ^ 3) / 8-b * p + a * p ^ 2 -p ^ 3-a * q + 2 * p * q = 0, db * q + q ^ 2 + a * p * qp ^ 2 * q = 0}, {p, q}); #a result: p = 1 / 2a; 4q ^ 2 + (a ^ 2-4b) q + 4d = 0 sample: p: = 1/2 * a; q: = (4 * b-a ^ 2 + sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; s: = 1/2 * a; t: = (4 * b-a ^ 2-sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; c: = (4 * a * b-a ^ 3) / 8; f: = x-> x ^ 4 + a * x ^ 3 + b * x ^ 2 + (4 * a * b-a ^ 3) / 8 * x + d; g: = x-> x ^ 2 + p * x + q; h: = x-> x ^ 2 + s * x + t; r: = x -> (cb * p + a * p ^ 2-p ^ 3-a * q + 2 * p * q) * x + (db * q + q ^ 2 + a * p * qp ^ 2 * q); simplify (r (x)); # Result: 0
A further simplification results with the relationships 3 4ab-a = 8c 4 2 2 2 - 2 - and a - 8ab + 16b - 64d = (a - 4b + 8 \ / d) · (a - 4b - 8 \ / d ) From this it follows: —————— / 2 —————— c / c 1/2 2 q = - + \ / —— - d = - (c + \ / c - ad) a 2 aa - ————— / 2 —————— c / c 1/2 2 t = - - \ / —— - d = - (c - \ / c - ada 2 aa Can be copied: q = c / a + sqrt (c ^ 2 / a ^ 2-d) t = c / a-sqrt (c ^ 2 / a ^ 2-d)

Here that Calculation scheme to calculate all zeros in Maple: f: = x-> x ^ 4 + a * x ^ 3 + b * x ^ 2 + 1/8 * (4 * a * b-a ^ 3) * x + d; p: = 1/2 * a; q: = (4 * b-a ^ 2 + sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; s: = 1/2 * a; t: = (4 * b-a ^ 2-sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8; g: = x -> (x ^ 2 + p * x + q) * (x ^ 2 + s * x + t); simplify (f (x) -g (x)); # sample is correct! x1: = - 1/4 * a + 1/4 * sqrt (3 * a ^ 2-2 * sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d) -8 * b); x2: = - 1/4 * a-1/4 * sqrt (3 * a ^ 2-2 * sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d) -8 * b); x3: = - 1/4 * a + 1/4 * sqrt (3 * a ^ 2 + 2 * sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d) -8 * b); x4: = - 1/4 * a-1/4 * sqrt (3 * a ^ 2 + 2 * sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d) -8 * b); h: = x -> (x-x1) * (x-x2) * (x-x3) * (x-x4); simplify (f (x) -h (x)); # sample is correct

### Example for the special case c = (4ab-a ^ 3) / 8 Short calculation

a = 4 b = 6 c = 4 // = (4ab-a ^ 3) / 8 d = 1 p = 1/2 * a = 2 s = 1/2 * a = 2 q = (4b-a ^ 2 + sqrt (a ^ 4-8a ^ 2 * b + 16b ^ 2-64d)) / 8 = 1 t = (4b-a ^ 2-sqrt (a ^ 4-8a ^ 2 * b + 16b ^ 2-64d )) / 8 = 1 x1 = -p / 2 + sqrt ((p / 2) ^ 2-q) = - 1 x2 = -p / 2-sqrt ((p / 2) ^ 2-q) = - 1 x3 = -s / 2 + sqrt ((s / 2) ^ 2-t) = - 1 x4 = -s / 2-sqrt ((s / 2) ^ 2-t) = - 1
b) f (x) = x4+ 2x3-11x2-12x + 36

a = 2 b = -11 c = -12 d = 36 p = 1/2 * a = 1 s = 1/2 * a = 1 q = (4 * ba ^ 2 + sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8 = -6 t = (4 * ba ^ 2-sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2- 64 * d)) / 8 = -6 x1 = -p / 2 + sqrt ((p / 2) ^ 2-q) = 2 x2 = -p / 2-sqrt ((p / 2) ^ 2-q) = -3 x3 = -s / 2 + sqrt ((s / 2) ^ 2-t) = 2 x4 = -s / 2-sqrt ((s / 2) ^ 2-t) = - 3
c) f (x) = x4+ 4x3-11x2-30x + 50

a = 4 b = -11 c = -30 d = 50 sample: c = (4 * a * ba ^ 3) / 8 = -30 q is the solution of the equation 4q ^ 2 + (a ^ 2-4b) q + 4d = 0 p = 1 / 2a = 2 s = 1 / 2a = 2 q = (4b-a ^ 2 + sqrt (a ^ 4-8a ^ 2 * b + 16b ^ 2-64d)) / 8 = -5 t = (4b-a ^ 2-sqrt (a ^ 4-8a ^ 2 * b + 16b ^ 2-64d))) / 8 = -10 x1 = -p / 2 + sqrt ((p / 2) ^ 2- q) = sqrt (6) -1 x2 = -p / 2-sqrt ((p / 2) ^ 2-q) = - 1-sqrt (6) x3 = -s / 2 + sqrt ((s / 2) ^ 2-t) = sqrt (11) -1 x4 = -s / 2-sqrt ((s / 2) ^ 2-t) = - 1-sqrt (11)

Calculation: a = 6 b = 9 c = 0 d = 1 p = 1/2 * a = 3 q = (4 * ba ^ 2 + sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2-64 * d)) / 8 = 1 * is = 1/2 * a = 3 t = (4 * ba ^ 2-sqrt (a ^ 4-8 * a ^ 2 * b + 16 * b ^ 2- 64 * d)) / 8 = -1 * i So f (x) = x ^ 4 + 6x ^ 3 + 9x ^ 2 + 1 = (x ^ 2 + 3x + i) (x ^ 2 + 3x- i)
You can see here: In the special case q is not always real (since q is the solution of a quadratic equation), but the decomposition succeeds anyway. And the zero can be calculated: x1 = -3 / 2-1 / 4 * sqrt (18 + 2 * sqrt (97)) - 1/4 * I * sqrt (-18 + 2 * sqrt (97)) x2 = -3 / 2 + 1/4 * sqrt (18 + 2 * sqrt (97)) + 1/4 * I * sqrt (-18 + 2 * sqrt (97)) x3 = -3 / 2-1 / 4 * sqrt (18 + 2 * sqrt (97)) + 1/4 * I * sqrt (-18 + 2 * sqrt (97)) x4 = -3 / 2 + 1/4 * sqrt (18 + 2 * sqrt (97 )) - 1/4 * I * sqrt (-18 + 2 * sqrt (97)) The decomposition into real quadratic factors is: (x-x1) (x-x3) = x ^ 2 + (3 + 1 / 2 * sqrt (18 + 2 * sqrt (97))) * x + 9/4 + 3/4 * sqrt (18 + 2 * sqrt (97)) + 1/4 * sqrt (97) and (x-x3 ) · (X-x4) = x ^ 2 + (3-1 / 2 * sqrt (18 + 2 * sqrt (97))) * x + 9 / 4-3 / 4 * sqrt (18 + 2 * sqrt ( 97)) + 1/4 * sqrt (97)

### II sub-case c = 0 and 4b = a2

We replace a with 2a and then come to the form:

### / ——————————————————————— \ | 4 3 2 2 | | x + 2ax + a x + d = 0 | (The coefficient in front of x is zero) \ ——————————————————————— /

Then results for the decomposition

### 4 3 2 2 2 2 x + 2ax + a x + d = (x + px + q) (x + sx + t)

—— —— with p = a, s = a, q = \ / - d, t = - \ / - d This can be recognized immediately by inserting and multiplying. Calculation scheme for Maple: p: = a; s: = a; q: = sqrt (-d); t: = - sqrt (-d); x1: = - p / 2 + sqrt ((p / 2) ^ 2-q); x2: = - p / 2-sqrt ((p / 2) ^ 2-q); x3: = - s / 2 + sqrt ((s / 2) ^ 2-t); x4: = - s / 2-sqrt ((s / 2) ^ 2-t);

#### Example 4 3 2 x - 6x + 9x - 3 = 0;

can be copied: x ^ 4-6 * x ^ 3 + 9 * x ^ 2-3 = 0 - - solution: p = -3, s = -3, q = \ / 3 t = - \ / 3 x1 = 3 / 2 + 1/2 * sqrt (9-4 * sqrt (3)) = 2.219686871 x2 = 3 / 2-1 / 2 * sqrt (9-4 * sqrt (3)) = 0.780313129 x3 = 3 / 2 + 1/2 * sqrt (9 + 4 * sqrt (3)) = 3.495507657 x4 = 3 / 2-1 / 2 * sqrt (9 + 4 * sqrt (3)) = - 0.495507657

### III special case p = s with further special conditions

(x ^ 2 + p * x + q) * (x ^ 2 + s * x + t) = x ^ 4 + a * x ^ 3 + (1/4 * a ^ 2 + t + q) * x ^ 2 + 1/2 * a * (t + q) * x + q * t Can I use the equation x4+ ax3+ bx2+ cx + d
divide the last coefficient d into d = q * t and it follows that
b = 1/4 * a ^ 2 + t + q and c = 1/2 * a * (t + q), then p = s = a / 2 and the decomposition is already known.

#### Example x4+ 4x3+ 12x2+ 16x + 15

For d = 15 = 3 · 5 one tries the solution q = 3 and t = 5. Sample:
b = 1/4 * a ^ 2 + t + q = 12
c = 1/2 * a * (t + q) = 16 Right (random)
Thus with p = s = a / 2
x4+ 4x3+ 12x2+ 16x + 15 = (x2+ 2x + 3) (x2+ 2x + 5)

### Decomposition: x4+ ax3+ bx2+ cx + d = (x2+ px + q) (x2+ sx + t)

This case can be recognized by the following relationship

### c = a sqrt (d)

We can then also use the equation with d = q2 in the following form.

### / ——————————————————————————— \ | 4 3 2 2 | | x + ax + bx + aqx + q = 0 | \ ——————————————————————————— /

Then it follows: a = p + s and b = 2q + ps
From this one can calculate p and s:
——————————— ———————————— 1 2 1 2 p = - (a + \ / a - 4b + 8q) s = - (a - \ / a - 4b + 8q) t = q 2 2
Sample with Maple: p: = 1/2 * a + 1/2 * sqrt (a ^ 2-4 * b + 8 * q); s: = 1/2 * a-1/2 * sqrt (a ^ 2-4 * b + 8 * q); yp: = expand ((x ^ 2 + p * x + q) * (x ^ 2 + s * x + q)); #Result yp = x ^ 4 + a * x ^ 3 + b * x ^ 2 + a * q * x + q ^ 2

#### Example x4+ 3x3-14x2+ 6x + 4 = 0

can be copied: x ^ 4 + 3 * x ^ 3-14 * x ^ 2 + 6 * x + 4 = 0
In this case a = 3, b = -14, c = 6 and d = 4 the condition c = a * sqrt (d) is fulfilled.
With q = t = 2 we get p = 6 s = -3
x1 = -3 + sqrt (7)
x2 = -3-sqrt (7)
x3 = 1
x4 = 2

#### Example x4+ 2x3+ 3x2+ 4x + 4 = 0

can be copied: x ^ 4 + 2 * x ^ 3 + 3 * x ^ 2 + 4 * x + 4 = 0
In this case a = 2, b = 3, c = 4 and d = 4 the condition c = a * sqrt (d) is fulfilled.
With q = t = 2 we get p and s to

p = 1 + sqrt (2) and s = 1-sqrt (2)
The four solutions to the equation x4+ 2x3+ 3x2+ 4x + 4 = 0 are therefore x1 = (- 1 / 2-1 / 2 * sqrt (2)) + (sqrt (5 / 4-1 / 2 * sqrt (2))) * I; x2 = (- 1 / 2-1 / 2 * sqrt (2)) - (sqrt (5 / 4-1 / 2 * sqrt (2))) * I; x3 = (1/2 * sqrt (2) -1/2) + (sqrt (1/2 * sqrt (2) +5/4)) * I; x4 = (1/2 * sqrt (2) -1/2) - (sqrt (1/2 * sqrt (2) +5/4)) * I; Maple cannot solve this equation.

### V sub-case: symmetric equation

A 4th degree equation of the following form is called symmetric. 4 3 2 x + ax + bx + ax + 1 = 0 This is a sub-case of special case IV and the following decomposition can be calculated immediately:

### 4 3 2 2 2 x + ax + bx + ax + 1 = (x + px + 1) (x + sx + 1)

——————————— —————————— 1 2 1 2 p = - (a + \ / a - 4b + 8) s = - (a - \ / a - 4b + 8) 2 2 can be copied: p = 1/2 * a + 1/2 * sqrt (a ^ 2-4 * b + 8) and s = 1/2 * a-1/2 * sqrt (a ^ 2-4 * b + 8)

#### Example x4+ 4x3-12x2+ 4x + 1 = 0

Can be copied: x ^ 4 + 4 * x ^ 3-12 * x ^ 2 + 4 * x + 1 = 0
This results in p = 2 + 3 * sqrt (2) q = 1, s == 2-3 * sqrt (2), t = 1 and thus x1 = -1-3 / 2 * sqrt (2) + 1 / 2 * sqrt ((2 + 3 * sqrt (2)) ^ 2-4) = - 0.164524665 x2 = -1-3 / 2 * sqrt (2) -1 / 2 * sqrt ((2 + 3 * sqrt (2)) ^ 2-4) = - 6.078116021 x3 = -1 + 3/2 * sqrt (2) + 1/2 * sqrt ((2-3 * sqrt (2)) ^ 2-4) = 1.628626278 x4 = -1 + 3/2 * sqrt (2) -1 / 2 * sqrt ((2-3 * sqrt (2)) ^ 2-4) = 0.6140144080

#### Example x4+ 2x3-14x2+ 2x + 1 = 0

can be copied: x ^ 4 + 2 * x ^ 3-14 * x ^ 2 + 2 * x + 1 = 0 p = 1 + sqrt (17) q = 1 s = 1-sqrt (17) t = 1 x1 = - 1 / 2-1 / 2 * sqrt (17) + 1/2 * sqrt (14 + 2 * sqrt (17)) = -0.203258342 x2 = -1 / 2-1 / 2 * sqrt (17) -1 / 2 * sqrt (14 + 2 * sqrt (17)) = -4.919847284 x3 = -1 / 2 + 1/2 * sqrt (17) + 1/2 * sqrt (14-2 * sqrt (17)) = 2.760905633 x4 = -1 / 2 + 1/2 * sqrt (17) -1 / 2 * sqrt (14-2 * sqrt (17)) = 0.362199993 (not solvable with Maple)

### VI sub-case: a = 0 and c = 0

A 4th degree equation of the following form is often called a biquadratic equation 4 2 x + bx + d = 0 This is a quadratic equation for x2 and the four solutions result immediately from the known formulas. ——————— ———— b / b 2 x = ± \ / w for w = - - ± \ / (-) - d 1,2,3,4 1,2 1,2 2 2 - ———— The computation of complex square roots \ / a + b · i = x + y · i is discussed in detail here.

The decomposition 4 2 2 2 x + bx + d = (x + px + q) (x + sx + t) then results from p = - (x + x), q = x x 1 2 1 2 s = - (x + x), t = x · x 3 4 3 4 where the zeros can be permuted as desired, which is the case with rellem b, d and desired real p, q, r and s becomes important.

#### Example x4+ 4x2+ 1 = 0 (b2 - 4d> 0)

With x1 = sqrt (-2 + sqrt (3)) = (1/2 * sqrt (6) - 1/2 * sqrt (2)) * i x2 = - sqrt (-2 + sqrt (3)) = - (1/2 * sqrt (6) - 1/2 * sqrt (2)) * i x3 = sqrt (-2-sqrt (3)) = (1/2 * sqrt (2) + 1/2 * sqrt ( 6)) * i x4 = - sqrt (-2-sqrt (3)) = - (1/2 * sqrt (2) + 1/2 * sqrt (6)) * i results with p = - (x1 + x2) etc. p = 0 q = 2-sqrt (3) s = 0 t = 2 + sqrt (3) i.e. x4+ 4x2+ 1 = (x2+ 2-√3) (x2+2+√3)

#### Example x4+ x2+ 4 = 0 (b2 - 4d <>

With x1 = 1/2 * sqrt (-2 + 2 * i * sqrt (15)) = 1/2 * sqrt (3) + 1/2 * sqrt (5) i x2 = -1 / 2 * sqrt (- 2 + 2 * i * sqrt (15)) = -1 / 2 * sqrt (3) - 1/2 * sqrt (5) i x3 = 1/2 * sqrt (-2-2 * i * sqrt (15) ) = 1/2 * sqrt (3) - 1/2 * sqrt (5) i x4 = -1 / 2 * sqrt (-2-2 * i * sqrt (15)) = -1 / 2 * sqrt (3 ) + 1/2 * sqrt (5) i results with p = - (x1 + x3) etc. p = -sqrt (3) q = 2 s = sqrt (3) t = 2 i.e. x4+ x2+ 4 = (x2-√3x + 2) (x2+ √3x + 2)

### Direct computation of real p, q, r and t from real b ≠ 0 and d ≠ 0.

x ^ 4 + b * x ^ 2 + d = (x ^ 2 + p * x + q) * (x ^ 2 + s * x + t) gives the system of equations: s + p = 0 t + p * s + q = bp * t + q * s = 0 q * t = d One can assume q ≠ 0 and immediately get the two conditions: t = d / q and s = -p and thus p * (dq ^ 2) = 0 and d / qp ^ 2 + q = b if d < 0="" oder="" 4*d="">< b^2="">< 0="" ergibt="" sich="" p="0," q="1/2*b+1/2*sqrt(b^2-4*d)," s="0," t="1/2*b-1/2*sqrt(b^2-4*d)" sonst="" ergibt="" sich="" q="sqrt(d)," p="sqrt(2*sqrt(d)-b)," s="-sqrt(2*sqrt(d)-b)," t="sqrt(d)" weiter="" fallunterscheidungen="" wegen="" des="" wurzelvorzeichens="" entfallen,="" da="" dann="" nur="" die="" werte="" von="" p,q="" und="" s="" und="" t="" vertauscht="" werden.="" beispiel="" a)="" b="1" d="4" ==""> p = sqrt (2 * sqrt (d) -b) = sqrt (3), q = sqrt (d) = 2, s = -sqrt (3), t = 2 Example b) b = 4 d = 1 => p = 0, q = 1/2 * b + 1/2 * sqrt (b ^ 2-4 * d) = 2 + sqrt (3), s = 0, t = 2-sqrt (3) example c) b = 1 d = -4 => p = 0, q = 1/2 * b + 1/2 * sqrt (b ^ 2-4 * d) = 1/2 + 1/2 * sqrt (17), s = 0, t = -1 / 2 * sqrt (17) +1/2 example d) b = 4 d = -1 => p = 0 q = 1/2 * b + 1/2 * sqrt (b ^ 2- 4 * d) = 2 + sqrt (5), s = 0, t = 3-sqrt (5)
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