#1**0 **

**Ok lets see...**

**alright lets make these look better.**

**\(\frac{x(x-4)}{(x-5)(x-4)}=\frac{4(x-5)}{(x-4)(x-5)}\)**

**alright more simplifying:**

**\(x(x-4)=4(x-5)\)**

**\(x^2-4x=4x-20\)**

**\(x^2-8x+20=0\)**

**Oh yay! A quadratic equation! Now time to solve!**

**Ok so:**

**\(x+y\:=\:-8,\:xy=20\)**

**\(x=-8-y\)**

**Substitute this into xy = 20 and you will get**

**\(\begin{pmatrix}x=-4+2i,\:&y=-4-2i\\ x=-4-2i,\:&y=-4+2i\end{pmatrix}\)**

**so x is NOT **4±2i **but it is -4\pm2i so yeah!**

**If anyone can check my work that would be great**

Imcool Dec 30, 2022

#2**0 **

Sorry, i tried typing in -4\pm2i and it said it was incorrect. Thank you tho for your effort in helping me!

Keihaku
Dec 30, 2022

#4**+1 **

We have:

x/x-5 = 4/x-4

Multiply by (x-5) and by (x-4) to both sides

x(x-4) = 4(x-5)

x^2 - 4x = 4x - 20

Put the left side on the right...

x^2 - 8x + 20 = 0

We find factors of 20 that multiply to 20, but add to -8 (we realize factors must be negative...)

Instead, we can just put this into our quadratic formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

So we have:

8 +/- root(64 - 80) }/2

8 +/- root(-16) }/2

8 +/- 4i

4 +/- 2i

So the answer **should **be **4** +/- 2i, note that the 4 is positive...

(If this answer is incorrect please notify me as I would be happy to help) (But you did say this is incorrect, so I don't know what's going wrong here...)

TooEasy Dec 30, 2022

#7**+1 **

\(x(x-4) = 4(x-5)\)

\(x^2 - 4x = 4x - 20\)

\(x^2 - 8x + 20 = 0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {8 \pm \sqrt{-16} \over 2}\)

\(x = {8 \pm 4i \over 2}\)

\(\color{brown}\boxed{x = 4 \pm 2i}\)

.BuilderBoi Dec 30, 2022