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# Find all values of x such that

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Find all values of x such that:

Answer is NOT 4±2i or 4\pm2i

Dec 30, 2022

#1
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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

Dec 30, 2022
#2
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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
#5
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@ImCool I'm trying to find your mistake gimme a sec...

YESSS I FOUND IT (I'm so pleased with my self)

Vieta states that with roots r and s in a simple quadratic:

r+s = negative b

I'm pretty sure this is right based on my knowledge, so that's where you went wrong I think...

TooEasy  Dec 30, 2022
edited by TooEasy  Dec 30, 2022
edited by TooEasy  Dec 30, 2022
#3
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The solutions are actually 3 \pm 2i.

Dec 30, 2022
#9
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That wasnt right either

Keihaku  Dec 30, 2022
#4
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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...)

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

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...)

Dec 30, 2022
edited by TooEasy  Dec 30, 2022
#6
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Honestly, I do think this is right, so could you try again a different way, or maybe they are asking for something more specific...

TooEasy  Dec 30, 2022
#8
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Strange... that's what I got as well, and Mathpapa verifies that

BuilderBoi  Dec 30, 2022
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$$x(x-4) = 4(x-5)$$

$$x^2 - 4x = 4x - 20$$

$$x^2 - 8x + 20 = 0$$

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

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

$$x = {8 \pm 4i \over 2}$$

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

.
Dec 30, 2022
#10
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Im not sure either

Keihaku  Dec 30, 2022
#11
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I think the problem had a typo.

x^2 - 4x = 4x + 20

it gives imaginary, perhaps the owner of the question meant the right hand side to be 5/(x - 4)

Dec 30, 2022
edited by proyaop  Dec 30, 2022