How do you solve #2/(x^2-16x+64)-8/(x-7)=(Ax+B)/(x^2-16x+63)#?

Question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-23T17:53:02

There is no solution.

#2/(x^2-16x+64)-8/(x-7)#

= #(2(x-7)-8(x^2-16x+64))/((x^2-2xx8xx x+8^2)(x-7))#

= #(2x-14-8x^2+128x-512)/((x-8)^2(x-7))#

= #(-8x^2+130x-526)/((x-8)^2(x-7))#

Further, #(Ax+B)/(x^2-16x+63)#

= #(Ax+B)/((x-7)(x-9))#, and hence

#(-8x^2+130x-526)/((x-8)^2(x-7))=(Ax+B)/((x-7)(x-9))#

or #((x-9)(-8x^2+130x-526))/((x-8)^2(x-7)(x-9))=((x-8)^2(Ax+B))/((x-8)^2(x-7)(x-9))#

or #(x-9)(-8x^2+130x-526)=(x^2-16x+64)(Ax+B)#

or #-8x^3+130x^2-526x+72x^2-1170x+4734=Ax^3-16Ax^2+64Ax+Bx^2-16Bx+64B#

or #-8x^3+202x^2-1696x+4734=Ax^3+(-16A+B)x^2+(64A-16B)x+64B#

Hence, comparing like terms, we should have

#A=-8#, #-16A+B=202#

now putting #A=-8# gives #B=74#

also then though #64A-16B=-512-1184=-1696#

and #64B=4736#

as the latter is not true

there is no solution.