If f(x) is a fifth degree polynomial, is it possible that it has exactly 5 complex roots?
1 answer:
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Answer:
FD≈25.94.. rounded = 26
Step-by-step explanation:
FD²=12²+(4x+11)²
FD²=144+16x²+88x+121
FD²=265+16x²+88x
also
FD²=12²+(13x-16)²
FD²=144+169x²-416x+256
FD²=400+169x²-416x
thus
265+16x²+88x = 400+169x²-416x
16x²-169x²+88x+416x+265-400 = 0
-153x²+504x-135 = 0
we will solve this quadratic equation by suing the quadratic formula to find x
x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)
x=(-504±)/2(-153)
x=(-504±)/-306
x=(-504±)/-306
x=(-504±414)/-306
x=(-504+414)/-306 and x=(-504-414)/-306
x=-90/-306 and x=-918/-306
x= 5/17 , 3
substituting x by the roots we found
check for 5/17:
4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..
13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..
check for 3:
4x+11 = 4×3+11 = 12+11 = 23
13x-16 = 13×3-16 = 23
thus 3 is the right root
therfore
ED=23 and CD=23
FD²=FE²+ED² or FD²=FC²+CD²
FD²=12²+23²
FD²=144+529
FD²=673
FD=√673
FD≈25.94.. rounded = 26