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Topic: Trigonometric-Algebraic and Hyperbolic-Algebraic Quadratic Equations
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Doctor Nisith Bairagi

Posts: 23
From: Uttarpara, West Bengal, India
Registered: 3/2/13
Trigonometric-Algebraic and Hyperbolic-Algebraic Quadratic Equations
Posted: Mar 28, 2013 11:04 AM
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From: Doctor Nisith Bairagi
Subject: Trigonometric-Algebraic and Hyperbolic-Algebraic Quadratic Equations
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 one-fourth 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 a-i 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 a-sinh 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). Trigonometric-algebraic quadratic equation
x^2-[cos a]x + 1 = 0,,,,,,,,,,Eq(2)

(2). Hyperbolic-algebraic 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
Trigonometric-algebraic, or, Hyperbolic-algebraic 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^2-1)/(na)]^2-[(4-n)/n]
To get, the square root of (sin a)^2, in the form: sin a = i(na^2-1)/(na), it is necessary that the second term, (4-n)/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: Trigonometric-Algebraic and Hyperbolic-Algebraic Quadratic equations.

Thanks to All.
Doctor Nisith Bairagi

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