Question

solve using quadratic formula
a^/(x-a)cube + 2a/(x-a)^ = 1/x - a
WILL FAN AND GIVE MEDAL

Answer 1

@johnweldon1993

Answer 2

Would you mind giving the equation in a better form?
it's a bit hard to understand.

Answer 3

wait a minute

Answer 4

a²/(x-a)³ + 2a/(x-a)² + 1/x -a

Answer 5

@Aditi_Singh

Answer 6

don not use the quadratic formula

Answer 7

just solve and explain it to me

Answer 8

Erm, You mean = 1/x-a.. na?

Answer 9

only for the last part

Answer 10

\[\frac{a^2}{(x−a)^3}+\frac{2a}{(x−a)^2}=\frac{1}{x−a}\]

For this equation I get the solution as : 2a ± √2a

Let me know if the answer is correct so that I could explain you..

Answer 11

Take \(\dfrac{1}{x-a} = k\) then solve the quadratic equation in variable \(a\).

Answer 12

@Aditi_Singh its + 1

Answer 13

i mean its not = to 1/x -a but its + 1/x-a

Answer 14

\[k^3 a^2 + 2k^2a + k=0\]Cancel \(k\) from both sides.\[k^2 a^2 + 2ka + 1=0\]By the quadratic formula\[a =\dfrac{-2k \pm\sqrt{4k^2 - 4k^2}}{2k^2} \]\[\Rightarrow a = \dfrac{-1}{(x-a)}\]

Answer 15

\[\dfrac{1}{x-a} = -a\]So\[a^2\cdot(-a)^3+2a\cdot(-a)^2 + (-a)=0\]