How many real solutions does this equation have?

Let f(x) = x^7 + 14 x^5 + 16 x^ 3 + 30 x - 560

f(1) = -499 and f(2) = 204 so there is a solution between 1 and 2.

f '(x) = 7x^6 + 70x^4 + 48x² + 30

This has only even powers and positive coefficients.

So f(x) is an increasing function.

Therefore the solution between 1 and 2 is the ONLY real solution.

Answer: 1

I don't see a way to factor that, and you can back up the rule of signs answer with calculus since the derivative of p(x) = x^7 + 14 x^5 + 16 x^ 3 + 30 x - 560 is:

p'(x) = 7x^6 + 70x^4 + 48x^2 + 30 .... is continuous and always positive

Since p(0) = -560,< 0, the sole root must be positive, and guess-and-check puts it between 1 and 2.

How to avoid Newton's method starting at x=2 (the steeper side, since p(x) is strictly icreasing) or so escapes me. I get the answer in 5 steps:

C:\Users\Husoski>k x=2

2

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.88603351955307

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.86968737279137

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.86938794009467

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.8693878416133

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.86938784161328

C:\Users\Husoski>k x = x - (x^^7 + 14*x^^5 + 16*x^^3 + 30*x - 560)/(7*x^^6 + 70*x^^4 + 48*x^^2 + 30)

1.86938784161328

The "k" command is my own homebrew calculator. The computations are done in IEEE double, until I get the extended precision ("bc killer") version done someday. The sixth step just verifies what was known from quadratic convergence of Newton's method...that final computation (if carried out with enough precision) is accurate to far more digits than are retained by IEEE double.

But, that's calculus, some numerical analysis and a touch of hocus pocus, not strictly algebra and definitely not quadratic. It'll be interesting to see what the Real Answer is.

This equation has one real solution.

x ≈ 1.86939

Note: Making x = 0 results in -560 ≠ 0

i) Applying Descartes's rule of signs,

When x is positive, f(x) sign is in the order, +, +, +, +, -

So there is only one change in sign

Hence it has a maximum of one real root.

ii) When x is negative, f(x) sign is in the order: -, -, -, -, -

So there is no change in the sign

Hence it has no real negative roots.

iii) Thus it can have only one real root;

Rest 6 must be complex roots

Also by the theory that complex roots occur only in pairs, and this being a polynomial of degree 7, it can have maximum 7 roots; of which complex roots has to be in pairs.

So it has one real positive root and 6 other complex roots.

Just one, the solution is 0.

Agreed on that answer. 0

x~~1.86939 real solution

x~~-1.66054-1.22254 i

x~~-1.66054+1.22254 i

x~~-0.129815-3.62409 i

x~~-0.129815+3.62409 i

x~~0.85566-2.1506 i

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