How do I show that the equation x^5 +2x^3 -x^2 -2x -6 only has one solution without using the graphical method

Question

How do I show that the equation x^5 +2x^3 -x^2 -2x -6 only has one solution without using the graphical method? Please explain

Answer 1

simple just find out DISCRIMINANT= b^2 - 4ac

if Dis is +ve then there are two real distinct roots

if Dis is 0 then there is exactly one root.

If the discriminant is negative, there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other

in your question

check for solving quintic equations and bring it in the form of
quadratic equation.

Answer 2

Interesting problem. I solved it as follows:

(1) There are no solutions for -1 < x < 1. To show this, note that if -1 < x < 1, then -1 < x^n < 1 for any integer n. Therefore, -1 < x^5 < 1, -1 < x^3 < 1, and 0 < x^2 <1. So, the quintic equation x^5 + 2x^3 - x^2 - 2x - 6 has to be less than 1 + 1*2 - 0*2 - (-2)*1 - 6 = 1 + 2 + 2 - 6 < 0. Therefore, this quintic is less than zero for all -1 < x < 1, proving it has no roots between -1 and 1.

(2) The equation is strictly increasing for x > 1. To show this, take the first derivative: f'(x) = 5x^4 + 6x^2 -2x -2. Since 6x^2 > 2x for all x > 1, f'(x) has to be at least 5 - 2, which is greater than zero, hence the quintic is increasing for all x > 1. Since the first step proved the quintic is less than zero for the range -1 < x < 1, we know there is exactly one root for x > 1.

(3) The equation is also strictly increasing for x < -1. Again, take the first derivative: f'(x) = 5x^4 + 6x^2 -2x -2. Since 6x^2 > 2x for all x < -1, f'(x) has to be at least 5 - 2, which is greater than zero, hence the quintic is increasing for all x < -1. Since the first step proved the quintic is less than zero for the range -1 < x < 1, we know there are no roots for x < -1.

qed!

Answer 3

the only way i will see to try this is utilising calculus. Take the spinoff and notice the place that is increasing and lowering. every time the spinoff is effective that is increasing and everywhere the spinoff is destructive that is lowering. everywhere it switches signs and indications the function has a community optimal or community minimum. evaluate the function on each of those durations and instruct that it in basic terms adjustments signs and indications one time. this might instruct that there is in basic terms one root to the function. there is not any rational root to this function, meaning that is an irrational root.

Answer 4

First- Any nth degree (here n=5) polynomial has at most n complex roots.

Second : One way would be to find all the roots
root1 = 1.395917950,
root2 = .1362911182+1.710032257*I,
root3 = -.8342500933+.8744321762*I,
root4 = -.8342500933-.8744321762*I, .
root5 = 1362911182-1.710032257*I
ie 4 complex roots, one real root. All 5 roots have been found and only one is real.