x 4 − 6 x 3 + 22 x 2 − 30 x + 13 = 0 x^4−6x^3+22x^2−30x+13=0 x 4 − 6 x 3 + 22 x 2 − 30 x + 13 = 0
Let f(x) = x4 −6x3 +22x2 −30x+13
f ( 1 ) = 1 − 6 + 22 − 30 + 13 = 0 f(1) = 1-6+22-30+13=0 f ( 1 ) = 1 − 6 + 22 − 30 + 13 = 0
So, ( x − 1 ) (x-1) ( x − 1 ) is a factor of f ( x ) . f(x). f ( x ) .
So, x 4 − 6 x 3 + 22 x 2 − 30 x + 13 = ( x − 1 ) x^4−6x^3+22x^2−30x+13 = (x-1) x 4 − 6 x 3 + 22 x 2 − 30 x + 13 = ( x − 1 ) (x 3 − 5 x 2 + 17 x − 13 x^3-5x^2+17x-13 x 3 − 5 x 2 + 17 x − 13 )
Let g ( x ) = x 3 − 5 x 2 + 17 x − 13 g(x)=x^3-5x^2+17x-13 g ( x ) = x 3 − 5 x 2 + 17 x − 13
g ( 1 ) = 1 − 5 + 17 − 13 = 0 g(1)=1-5+17-13=0 g ( 1 ) = 1 − 5 + 17 − 13 = 0
So, ( x − 1 ) (x-1) ( x − 1 ) is a factor of g ( x ) . g(x). g ( x ) .
Thus, f ( x ) = ( x − 1 ) . g ( x ) = ( x − 1 ) ( x − 1 ) ( x 2 − 4 x + 13 ) f(x)=(x-1).g(x)=(x-1)(x-1)(x^2-4x+13) f ( x ) = ( x − 1 ) . g ( x ) = ( x − 1 ) ( x − 1 ) ( x 2 − 4 x + 13 )
Now, put ( x 2 − 4 x + 13 ) = 0 (x^2-4x+13)=0 ( x 2 − 4 x + 13 ) = 0
Using Quadratic formula:
x = 4 ± 16 − 52 2 = 4 ± 6 i 2 = 2 ± 3 i x=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3i x = 2 4 ± 16 − 52 = 2 4 ± 6 i = 2 ± 3 i
Hence, the solution are x = 1 , 1 , 2 + 3 i , 2 − 3 i x=1,1,2+3i,2-3i x = 1 , 1 , 2 + 3 i , 2 − 3 i and rational solutions are x = 1 , 1 x=1,1 x = 1 , 1
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