Fraction of the Year

Algebra Level 3

1 x + 2 + 7 x + 0 + 5 x + 1 + 3 x + 8 = 0 \frac1{x+2}+\frac7{x+0}+\frac{-5}{x+1}+\frac3{x+8}=0 Find x |x| if x x is real.


The answer is 3.5.

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1 solution

Naren Bhandari
May 14, 2018

1 x + 2 + 7 x 5 x + 1 + 3 x + 8 = 0 8 x + 14 x ( x + 2 ) 2 x + 37 ( x + 1 ) ( x + 8 ) = 0 \frac{1}{x+2} +\frac{7}{x} -\frac{5}{x+1} +\frac{3}{x+8} =0 \\ \frac{8x+14}{x(x+2)} -\frac{2x+37}{(x+1)(x+8)} =0 On further simplication of last equation we get 6 x 3 + 45 x 2 + 116 x + 112 = 0 ( 2 x + 7 ) ( 3 x 2 + 12 x + 16 ) = 0 2 x + 7 = 0 3 x 2 + 12 x + 16 = 0 6x^3 +45x^2 +116x+112 =0 \\ (2x+7) (3x^2 +12x+16) =0 \\ 2x+7 =0 \qquad 3x^2 +12x+16 =0 One of the value we have x = 7 2 x= -\frac{7}{2} , the discriminant of quadratic equation is D < 0 D <0 , which gives no real solution in x x .

Only real solution of x x is 7 2 -\frac{7}{2} . Therefore, x = 3.5 |x| =3.5


It will be easier if you add it up like this ( 1 x + 2 + 3 x + 8 ) + ( 7 x + 0 + 5 x + 1 ) (\frac1{x+2}+\frac3{x+8})+(\frac7{x+0}+\frac{-5}{x+1})

X X - 3 years ago

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