#x^7+x^5 = 81# ?
1 Answer
This septic equation has exactly one real root:
#x ~~ 1.80302#
Explanation:
Given:
#x^7+x^5 = 81#
Subtract
#x^7+x^5-81 = 0#
By the rational roots theorem, any rational roots of this septic equation are expressible in the form
That means that the only possible rational roots are:
#+-1, +-3, +-9, +-27, +-81#
Note that the signs of the coefficients are in the pattern
Reversing the signs on the terms of odd degree, we get the pattern
We find:
#(color(blue)(1))^7+(color(blue)(1))^5-81 = 1+1-81 = -79 < 0#
#(color(blue)(3))^7+(color(blue)(3))^5-81 = 2187+243-81 = 2349 > 0#
So the equation has no rational roots. It has exactly one positive irrational real root in
In common with most polynomials of degree
About the best we can do is find numerical approximations such as:
#x ~~ 1.80302#
graph{x^7+x^5-81 [-5, 5, -1000, 1000]}
To find approximations to the real root, you would probably use Newton's method. To get all
For Newton's method, define:
#f(x) = x^7+x^5-81#
Then:
#f'(x) = 7x^6+5x^4#
Given an approximation
#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (a_i^7+a_i^5-81)/(7a_i^6+5a_i^4)#
Putting this formula into a spreadsheet and setting
#a_0 = 2#
#a_1 ~~ 1.85037878787879#
#a_2 ~~ 1.80631325284238#
#a_3 ~~ 1.80303898643730#
#a_4 ~~ 1.80302212305041#
#a_5 ~~ 1.80302212260564#
#a_6 ~~ 1.80302212260564#