Aug 14, 2020 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,812 Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.
Aug 14, 2020 #2 kaliprasad Well-known member Mar 31, 2013 1,334 Spoiler: My solution We have $x^8-x^7 + x^2 -x + 15 = x^7(x-1) + x(x-1) + 15$ each term is positive for $x > 1$ so LHS is greater than 0 so no solution for $ x > 1$ for x = 1 LHS = 15 so x = 1 is not a solution Further $x^8-x^7 + x^2 -x + 15 = (15- x) + x^2(1-x^5) + x^8 $ Each term is positive for $x < 1$ so LHS is greater than 0 so no solution for $ x < 1$ Hence no real solution
Spoiler: My solution We have $x^8-x^7 + x^2 -x + 15 = x^7(x-1) + x(x-1) + 15$ each term is positive for $x > 1$ so LHS is greater than 0 so no solution for $ x > 1$ for x = 1 LHS = 15 so x = 1 is not a solution Further $x^8-x^7 + x^2 -x + 15 = (15- x) + x^2(1-x^5) + x^8 $ Each term is positive for $x < 1$ so LHS is greater than 0 so no solution for $ x < 1$ Hence no real solution