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Given polynomial  equation : -x⁴ + 80 x + 36 = 0

Let us express an equation of degree 4 as
          P(x) =  (x - a)(x-b)(x - c) ( x- d) = 0

Thus either all are real roots, or two of them are real or none of the roots are real. the reason is that we know that imaginary roots appear in pairs.

expand P(x) = [ x² - (a+b) x + ab ] [ x² - (c+d) x + cd ] = 0
    x⁴ - [a+b+c+d] x³ + x² [ ab+cd+ac+ad+bc+bd] - [abc+abd+acd+bcd] x +abcd=0

Sum of roots a+b+c+d = 0 as in the given equation the coefficient of x³ term is 0.


if the given equation is a quadratic equation :  perhaps there is a small typing mistake in that.  then

  Q(x) =  -x² + 80 x + 36  = 0 
  let P(x) =   x²  - 80 x - 36 = 0
        Using the change of sign of the coefficients in P(x) we know that  there is one positive real root and one negative real root.
       the sum is the minus of coefficient of the x term.    80.

    quadratic equation is expressed as (x - a)(x - b) = x² - (a+b)x+ab = 0

2 5 2
hope it is easy to understand
sir what about my doubts
you please do it sir
Thanks kvn murthy garu
you know telugu?
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