Solve #(x+3)/(x+2) + (x+4)/(x+3) = (x+5)/(x+4) - (x+6)/(x+5)#?
2 Answers
The solution of:
#(x+3)/(x+2)color(red)(-)(x+4)/(x+3) = (x+5)/(x+4)-(x+6)/(x+5)#
is
Explanation:
Suppose the question should be:
#(x+3)/(x+2)color(red)(-)(x+4)/(x+3) = (x+5)/(x+4)-(x+6)/(x+5)#
Making common denominators on the left hand side and on the right hand side, this becomes:
#((x+3)(x+3)-(x+2)(x+4))/((x+2)(x+3)) = ((x+5)(x+5)-(x+4)(x+6))/((x+4)(x+5))#
Multiplying out the numerators, we get:
#((x^2+6x+9)-(x^2+6x+8))/((x+2)(x+3)) = ((x^2+10x+25)-(x^2+10x+24))/((x+4)(x+5))#
Most of the terms in the numerator cancel, to give us:
#1/((x+2)(x+3)) = 1/((x+4)(x+5))#
Taking the reciprocal of both sides, this becomes:
#(x+2)(x+3) = (x+4)(x+5)#
which multiplies out as:
#x^2+5x+6 = x^2+9x+20#
Subtracting
#-14 = 4x#
Dividing both sides by
#x = -7/2#
In the form given this resolves to a typical quartic with approximate roots:
#x_1 ~~ -9.4400#
#x_2 ~~ -0.28158#
#x_3 ~~ -2.6392+4.5893i#
#x_4 ~~ -2.6392-4.5893i#
Explanation:
Assuming the question is correct as given...
Given:
#(x+3)/(x+2)+(x+4)/(x+3) = (x+5)/(x+4)-(x+6)/(x+5)#
Subtract the right hand side from the left to get:
#(x+3)/(x+2)+(x+4)/(x+3)-(x+5)/(x+4)+(x+6)/(x+5) = 0#
Transposing and multiplying both sides by
#0 = (x+3)^2(x+4)(x+5)+(x+2)(x+4)^2(x+5)-(x+2)(x+3)(x+5)^2+(x+2)(x+3)(x+4)(x+6)#
#color(white)(0) = (x^4+15x^3+83x^2+201x+180)+(x^4 +15x^3+82x^2+192x+160)-(x^4+15x^3+81x^2+185+150)+(x^4+15x^3+80x^2+180x+144)#
#color(white)(0) = 2x^4+30x^3+164x^2+573x+149#
This is a typical quartic, with two real irrational zeros and two non-real complex zeros.
It is possible but very messy to solve algebraically. Using a numerical method such as Durand-Kerner we find approximate solutions:
#x_1 ~~ -9.4400#
#x_2 ~~ -0.28158#
#x_3 ~~ -2.6392+4.5893i#
#x_4 ~~ -2.6392-4.5893i#
See https://socratic.org/s/aKtpkf7J for more details.
