
TrigonometricAlgebraic and HyperbolicAlgebraic Quadratic Equations
Posted:
Mar 28, 2013 11:04 AM


From: Doctor Nisith Bairagi Subject: TrigonometricAlgebraic and HyperbolicAlgebraic Quadratic Equations (sci.math.independent) Date: March 28, 2013. ??????????????????????????????????. If the pair of roots [a, 1/(4a)], of a quadratic equation in x, is such that, one of the roots is onefourth of the reciprocal of the other, the quadratic equation is:
x^2[a +1/(4a)]x + 1 = 0 or, x^2[(4a^2 + 1)/(4a)]x + 1 = 0,,,,,,,,,,Eq(1) or, 4ax^2(4a^2 + 1)x + 1/4 = 0 Rewrite this equation replacing [a +1/(4a)], either with: cos a, or, with: cosh(a):
Let, cos a = a +1/(4a) = (4a^2 + 1)/(4a) sin a = i(4a^2 + 1)/(4a) e^(ia) = cos a + i sin a = 1/(2a), e^(ia) = cos ai sin a = 2a e^(ia) + e^(ia) = 2 cos a = (4a^2 + 1)/(2a) This confirms that: cos a = (4a^2 + 1)/(4a)
Again, replacing (ia) by, i(ia), e^(a) = cos(ia) + i sin(ia) = cosh asinh a = 1/(2a) e^(a) = cos(ia)i sin(ia) = cosh a + sinh a = (2a) [e^(a) + e^(a)]/2 = cosh a = (4a^2 + 1)/(4a) This confirms that: cosh a = (4a^2 + 1)/(4a)
Note that, for each of the substitution for (cos a) or, (cosh a) for (4a^2 + 1)/(4a), the result is the same quadratic equation, viz., Eq(1).
Thus, the given quadratic equation Eq(1), having roots [a, 1/(4a)], namely: x^2[(4a^2 + 1)/(4a)]x + 1 = 0, has the alternative equivalent forms as:
(1). Trigonometricalgebraic quadratic equation x^2[cos a]x + 1 = 0,,,,,,,,,,Eq(2)
(2). Hyperbolicalgebraic quadratic equation x^2[cosh a]x + 1 = 0,,,,,,,,,,Eq(3)
That, only for n = 4, the quadratic equation with the pair of roots [a, 1/(na)], yields Trigonometricalgebraic, or, Hyperbolicalgebraic quadratic equations, will be imminent.
The quadratic equation is: x^2[a +1/(na)]x + 1/n = 0 or, x^2[(na^2 + 1)/(na)]x + 1/n = 0 or, nax^2(na^2 + 1)x + a = 0
Rewrite this equation replacing: cos a = a +1/(na) = (na^2 + 1)/(na) from which, (sin a)^2 = 1(cos a)^2 = [(na^21)/(na)]^2[(4n)/n] To get, the square root of (sin a)^2, in the form: sin a = i(na^21)/(na), it is necessary that the second term, (4n)/n, should be = 0, which demands that: n = 4.
Similar criterion (i.e., n = 4), is also applicable for the quadratic equation with, cosh a = a +1/(na) = (na^2 + 1)/(na). ??????????????????????????????????.. Dear Readers, Please comment on my observation and on the naming of the quadratic equations as: TrigonometricAlgebraic and HyperbolicAlgebraic Quadratic equations.
Thanks to All. Doctor Nisith Bairagi ?????????????????????????????????.

