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[SOLVED] Isolate y in terms of x

Bmanmcfly

Member
Mar 10, 2013
42
I feel like I should know this already, but I need to isolate y in the equation:
\(\displaystyle x= 6y-y^2\)
I was thinking of using the quadratic formula, but that would keep the equation in terms of x.

No matter how I seem to slice it, I can't figure out what to do... Which is kinda sad that I'm doing good on the larger problems, but struggle with what seems like should be simple.

Thanks for any guidance.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
I feel like I should know this already, but I need to isolate y in the equation:
\(\displaystyle x= 6y-y^2\)
I was thinking of using the quadratic formula, but that would keep the equation in terms of x.

No matter how I seem to slice it, I can't figure out what to do... Which is kinda sad that I'm doing good on the larger problems, but struggle with what seems like should be simple.

Thanks for any guidance.
Welcome to MHB, Bmanmcfly! :)

You are right to use the quadratic formula.
To do so, you need to rewrite it to the proper form.
And yes, in the result you will still have an x.
That is okay, you should treat x in this case just like a regular number.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I feel like I should know this already, but I need to isolate y in the equation:
\(\displaystyle x= 6y-y^2\)
I was thinking of using the quadratic formula, but that would keep the equation in terms of x.

No matter how I seem to slice it, I can't figure out what to do... Which is kinda sad that I'm doing good on the larger problems, but struggle with what seems like should be simple.

Thanks for any guidance.
You have to solve the second order equation $\displaystyle y^{2} - 6\ y + x=0$ where y is the unknown. Note that the (1) in general has two solutions that may be also complex...

Kind regards

$\chi$ $\sigma$
 
Last edited:

Bmanmcfly

Member
Mar 10, 2013
42
Is this correct?? I used quadratic formula, and it gave me:

Y=3+/-sqrt(36-4x)

Or should I try this by "completing the square"?

Edit: This gave me y= 3+/-sqrt(9-x),

So, it seems either way I'm doing something right and/or something wrong
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Is this correct?? I used quadratic formula, and it gave me:

Y=3+/-sqrt(36-4x)

Or should I try this by "completing the square"?
Either way works.
Your result is almost correct.
But it seems you made a mistake in the process.

You should have:
$$y=\frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot x}}{2 \cdot 1}$$
 

Bmanmcfly

Member
Mar 10, 2013
42
[solved]Re: Isolate y in terms of x

Either way works.
Your result is almost correct.
But it seems you made a mistake in the process.

You should have:
$$y=\frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot x}}{2 \cdot 1}$$
Thanks, ya, you're right. I feel kinda dumb to be having this as an issue.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
The method of completing the square fits amazingly well here:

$$-y^2 +6y = x \implies -y^2 +6y -9 = x-9 \implies (3-y)^2 = x-9,$$

and from that we have that $y = 3 \pm \sqrt{x-9}$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The method of completing the square fits amazingly well here:

$$-y^2 +6y = x \implies -y^2 +6y -9 = x-9 \implies (3-y)^2 = x-9,$$

and from that we have that $y = 3 \pm \sqrt{x-9}$.
One small quibble...you want:

\(\displaystyle (3-y)^2=9-x\)
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I didn't understand. :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle -y^2+6y-9=-(3-y)^2\)