from 4:00 a(a + 6)(a + 3)(a - 3) = -45, let t = a + 1.5(which is mean value of 4 terms) then (t - 1.5)(t + 4.5)(t + 1.5)(t - 4.5) = -45 => (t^2 - 1.5^2)(t^2 - 4.5^2) + 45 = 0 => 16t^4 - 360t^2 + 1449 = (4t^2 - 21)(4t^2 - 69) = 0 t^2 = 21/4 or 69/4 => we can get t , a = t - 1.5 and x = a^2
Setting t = √x, we get t^4+6t^3-9t^2-54t+45=0. Let us try (t^2+at-3)(t^2-bt-15) = 0> a+b=6 and ab=9 > a=b=3. The quartic equation factored out into two quadratics, t^2+3t-15=0, t^2+3t-3=0 The roots are t= 1/2[-3 +/-√69], t= 1/2[-3 +/-√21]. Of these, only t=1/2[-3+√69], 1/2[-3+√21] are valid. So, x=1/2[39-3√69], 1/2[15-3√21].
Let √x = t, (x , t > 0) (*). The given equation equivalent with the equation, after some algebra, t^4 + 6 t^3 - 9 t^2 - 54 t + 45 =0 or (t^2 + 3 t - 3) • (t^2 + 3t - 15) = 0 (#). The t^2 + 3 t - 3 =0 have roots t = (- 3 ± √21)/2 (**). The t^2 + 3 t - 15 = 0 have roots t = (- 3 ± √69)/2 (***). From (*), (**), (***) only the roots t = (-3 + √21)/2 and t = (-3 + √69)/2 are accepted. From (*) x = t^2 => x = [(-3 + √21)/2]^2 = (15 - 3√21)/2 and x = [(-3 + √69)/2]^2 = (38 - 3√69)/2 .. Note (#). (Old classic method). t^4 + 6 t^3 - 9 t^2 - 54 t +45 = ( t^2 + a • t + b) • (t^2 + c•t+ d) => a + c = 6 b + d + a • c = - 9 a • d + b • c = - 54 b • d = 45 .. ( a, b, c, d integers) .. Because b • d = 45 => b • d = ( - 1)(-45) = (-3)(-15) = (-5)(-9) = 1 • 45 = 3 • 15 = 5 • 9 ... We easily observe that ( b, d) = (- 3, -15) and it follows that ( a , c) = ( 3, 3) .
amazing
from 4:00 a(a + 6)(a + 3)(a - 3) = -45, let t = a + 1.5(which is mean value of 4 terms)
then (t - 1.5)(t + 4.5)(t + 1.5)(t - 4.5) = -45 => (t^2 - 1.5^2)(t^2 - 4.5^2) + 45 = 0
=> 16t^4 - 360t^2 + 1449 = (4t^2 - 21)(4t^2 - 69) = 0
t^2 = 21/4 or 69/4 => we can get t , a = t - 1.5 and x = a^2
Setting t = √x, we get t^4+6t^3-9t^2-54t+45=0. Let us try (t^2+at-3)(t^2-bt-15) = 0> a+b=6 and ab=9 > a=b=3. The quartic equation factored out into two quadratics, t^2+3t-15=0, t^2+3t-3=0 The roots are t= 1/2[-3 +/-√69], t= 1/2[-3 +/-√21]. Of these, only t=1/2[-3+√69], 1/2[-3+√21] are valid. So, x=1/2[39-3√69], 1/2[15-3√21].
My answer is right.
X=78/9 and root X=9
(XSqrt[X])/9+5/(Sqrt[X]+6)=Sqrt[X] X=(15-3Sqrt[21] )/2=7.5-1.5Sqrt[21] X=(39-3Sqrt[69])/2=19.5-1.5Sqrt[69]
Let √x = t, (x , t > 0) (*).
The given equation equivalent with the equation, after some algebra,
t^4 + 6 t^3 - 9 t^2 - 54 t + 45 =0 or
(t^2 + 3 t - 3) • (t^2 + 3t - 15) = 0 (#).
The t^2 + 3 t - 3 =0 have roots
t = (- 3 ± √21)/2 (**).
The t^2 + 3 t - 15 = 0 have roots
t = (- 3 ± √69)/2 (***).
From (*), (**), (***) only the roots
t = (-3 + √21)/2 and t = (-3 + √69)/2
are accepted.
From (*)
x = t^2 =>
x = [(-3 + √21)/2]^2 = (15 - 3√21)/2
and
x = [(-3 + √69)/2]^2 = (38 - 3√69)/2 ..
Note (#). (Old classic method).
t^4 + 6 t^3 - 9 t^2 - 54 t +45 =
( t^2 + a • t + b) • (t^2 + c•t+ d) =>
a + c = 6
b + d + a • c = - 9
a • d + b • c = - 54
b • d = 45 .. ( a, b, c, d integers) ..
Because b • d = 45 =>
b • d = ( - 1)(-45) = (-3)(-15) = (-5)(-9) = 1 • 45 = 3 • 15 = 5 • 9 ...
We easily observe that
( b, d) = (- 3, -15) and it follows that
( a , c) = ( 3, 3) .
{x+x ➖ }{x +x ➖}/9+9 ➖ +{5+5 ➖ }/{x+x ➖ 6+6 ➖ }= {x^2+x^2}/18+10/{x^2+12}= {x^4/18+10/12x^2 }= 10x^4/30x^2 2^5x^2^2/12^18x^2 1^1x^1^1/12^9^9x^1 /3^43^2^3^2 /1^1^2^1^1^3^1 /23 (x ➖ 3x+2).