Help needed finding maximum number of real solutions of the equation.?

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Equation is

ax^n-bx^2+cx-d=0

if n is positive even(n<2).

3

✅ Answers

? Favorite Answer

  • The simplest mathematical way would be to differentiate…

    dy/dx=nax^(n-1) -2bx

    Max or min

    0=nax^(n-1) -2bx

    0=x(nax^(n-2)) -2b)

    x=0, and nax^(n-2) =2b are solutions

    x^(n-2) =2b/na

    for real solutions. 2b/na must be positive

    This produces 2 solutions ±(n-2) root(2b/na)

    So 3 max and mins

    so 4 real solutions

  • 2ªn + 2ªX

    2 × 2 / 2.2 = 1.81

    1.81ªX X / 181 = 0.90

    0.90ªX X / 0.90 = 0.502

    Source(s): X = 0.5

  • n

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