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

Question

1707 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.

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