Question

evaluate :x4-x3+X2-x+1for x=2

Answer 1

f(x)=x⁴-x³+x²-x+1
f(2)=(2)⁴-(2)³+(2)²-2+1
=16-8+4-2+1
=21-10 =11...
so, 11 is the right answer for x=2

Answer 2

x^4+x^3+x^2+x+1 = 0.

Divide by x^2.

x^2+x+1+1/x+1/x^2 = 0.

(x^2+1/x^2) +(x+1/x) +1 = 0

(x+1/x)^2 -2 + (x+1/x)+1 =0

(x+1/x)^2 +(x+1/x) -1 = 0. This is aquadratic equation in (x+1/x). Or

y^2 +y -1 =  0 , where y = x+1/x.

y1 = (-1+sqrt5)/2 , or y2 = (-1-sqrt5)/2

y1 = x+1/x =  (-1+sqrt5)/2

x^2 +(1-sqrt5)/2 *x  +1 = 0

x1  =  {(-1+sqrt5)/2 + sqrt[(1-sqrt5)^2/4 -4}/2

x1 = {-(1-sqrt5)/4 + sqrt(-5-sqrt5)/2}/2

x1 = {-(1-sqrt5)/4 +sqrt(-1)sqrt(5+sqrt5)/4}

x2 = {-(1-sqrt5)/4 -sqrt(-1) sqrt(5+sqrt5)/4

Similarly we can solve for y2 = x+1/x = (-1-srqt5)/2

x^2 +(1+sqrt5)x+1 = 0

x3 = -(1+sqrt5) /4+ [sqrt(sqrt5 -5)]/4 or

x3  = -(1+sqrt5)/4 +{sqrt(-1)sqrt(5-sqrt5)}/4

x4 = -(1+sqrt5)- {sqrt(-1) sqrt(5-sqrt5)}/4

S0 there are as above 4 complex roots x1,x,2 ,x3 and x4..


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