Need assistance in solving this equation

[Cafelatte]

Hi! I need some help in solving this equation. I don't know where to start!

Let k ​be a constant.


So I know that we're going to times x+k to (x-k)x so that it will have the same denominator as the rest. After that I have no idea what to do next!

 

[HallsofIvy]

Hi! I need some help in solving this equation. I don't know where to start!

Let k ​be a constant.
View attachment 4919

So I know that we're going to times x+k to (x-k)x so that it will have the same denominator as the rest. After that I have no idea what to do next!

Rather than get common denominators, I would just multiply the entire equation by x+ k. That will give (x+ k)(x- k)x as the first term and get rid of the denominators in the others. Multiply each term out. You will get a cubic equation but at least one root is fairly easy to get "by inspection". Use the quadratic formula to get the other two.

 

[pka]

Hi! I need some help in solving this equation. I don't know where to start!

Let k ​be a constant.
View attachment 4919!

Write it as \(\displaystyle \Large{(x^2-k^2)x+kx(x-k)=-k^2(x-k)}\)

 

[Ishuda]

Hi! I need some help in solving this equation. I don't know where to start!

Let k ​be a constant.
View attachment 4919

So I know that we're going to times x+k to (x-k)x so that it will have the same denominator as the rest. After that I have no idea what to do next!

Am I not seeing something? (x-k) is a common factor, so one solution is x = k. For other solutions, assume x is not equal to k and cancel across. Then, assuming x is not equal to minus k, multiply through by x+k to get
x (x+k) + kx = -k²
or
x² + 2 k x + k² = (x + k)² = 0
or
x=-k which contradicts the assumption we made to get here that x was not minus k. So the only solution is x=k.

EDIT: Whoops, didn't notice was almost the same as Denise's so guess I wasn't missing something.

 

[lookagain]

This is just my way...disregard if you don't like!
I find it much faster, plus much clearer, plus makes cancellations easier.

Let a = x - k and b = x + k
So equation becomes:
ax + akx / b = -ak^2 / b

Multiply by b:
abx +akx = -ak^2

Divide by a:
bx + kx = -k^2 \(\displaystyle \ \ \ \ \ \ \) Denis, you lost information when you divided by a. It contains the variable x.
You can't reduce the equation from a 3rd degree to a 2nd degree without accounting for one of the solutions.
That solution is x = k.


Substitute b back in:
x(x + k) + kx = -k^2

Simplify:
x^2 + 2kx + k^2 = 0

Finish it...

Be aware in finishing it that you must check the candidate solution(s)
against restrictions of the original equation.

 

[Cafelatte]

Thank you! I've manged to figure how to solve the equation with all your help.

I found it easier by multiplying the (x+k) term first (as suggested by HallsofIvy) and continuing the step with taking the common factor (x-k) from the three terms, making it (x-k)(k²+2kx+x²) - thank you Ishuda for this suggestion.

And I'd also like to thank Denis, your method will surely help me in other Math equations that I'll come across in the future. It'll definitely save me the time and like you said, it's clearer and easier!

 

[Deleted member 4993]

Thank you! I've manged to figure how to solve the equation with all your help.

I found it easier by multiplying the (x+k) term first (as suggested by HallsofIvy)

Be careful at this step! You are assuming \(\displaystyle (x+k) \ \ne 0 \ \ i.e. \ x \ \ne \ -k\)

and continuing the step with taking the common factor (x-k) from the three terms, making it (x-k)(k²+2kx+x²) - thank you Ishuda for this suggestion.

And I'd also like to thank Denis, your method will surely help me in other Math equations that I'll come across in the future. It'll definitely save me the time and like you said, it's clearer and easier!

.

 

[Ishuda]

...
I found it easier by multiplying the (x+k) term first ...

Just to add to what Subhotosh Khan mentioned above [the "Be careful ..."]. Although we can write the original expression as
\(\displaystyle \frac{(x-1) (x+k)^2}{x+k}\)
and then tacitly cancel the x+k to get (x-1)(x+k), what we have done is equivalent to removing the discontinuity of a function. That is
f(x) = \(\displaystyle (x-k)\, x + \frac{k\, x\, (x-k)}{x+k} + \frac{k^2\, (x-k)}{x+k}\)
is defined everywhere but at x = -k and thus is not continuous at x=-k. However, we can define f(-k) to be zero and make the function not only continuous but, in this case, differentiable at k=-x.