Factoring with parameter

[Kalen12]

So here is the problem: xm²-4x²=m-1-3x
I need to make this equation in the form of: a*x=b
Farthest I've got is: (4x+1)(1-x)+m(xm-1)=0; but this doesn't really lead me to the solution ( or I don't see it).
Would appreciate help, thanks in advance.

 

[MarkFL]

I would write the equation in the form:

[MATH]4x^2-(m^2+3)x+(m-1)=0[/MATH]
Now you can apply the quadratic formula.

 

[Kalen12]

I would write the equation in the form:

[MATH]4x^2-(m^2+3)x+(m-1)=0[/MATH]
Now you can apply the quadratic formula.

You misinterpeted my question. I do not need to find a solution for x. Using the quadratic formula I would just get some x₁ and x₂, but what I'm really looking for is the "standard" form of a*x = b .
Here's an example:
m²*x -1= x+m;
x(m²-1)=m+1;

thus: a = (m²-1) and b=m+1

I would need something like this ( if plausible at all!) but thanks for the help, using the quadratic might help but I don't see it but will sure try though.
I hope you've understood me. Again, it's not the solution for x I seek which can be expressed with the quadratic formula for the parameter m, but this "form" so to speak of a*x=b.

 

[MarkFL]

You can't turn a quadratic equation into a linear one, unless you are approximating over a small interval I suppose. In the example you posted there are no terms with \(x^2\) as a factor, but in the orginal equation there is.

 

[Dr.Peterson]

So here is the problem: xm²-4x²=m-1-3x
I need to make this equation in the form of: a*x=b
Farthest I've got is: (4x+1)(1-x)+m(xm-1)=0; but this doesn't really lead me to the solution ( or I don't see it).
Would appreciate help, thanks in advance.

Since the original equation is quadratic in x, it is impossible to rewrite it as a constant times x equals a constant. Your other example doesn't have x squared, so it is not comparable.

The best you could do to factor your equation would be to write it as something like a(x-b)(x-c) = 0; and then b and c will be the two roots found by the quadratic formula.

 

[Kalen12]

You can't turn a quadratic equation into a linear one, unless you are approximating over a small interval I suppose. In the example you posted there are no terms with \(x^2\) as a factor, but in the orginal equation there is.

Well then no need to go further, I see the it now. I kind of forgot about that, thought there was some clue perhaps since I do not know all the conditions of the problem but now it's been clarified and thank you very much, sir!
I was trying to evade the quadratic formula since I saw it lead to an "ugly" expression.

Since the original equation is quadratic in x, it is impossible to rewrite it as a constant times x equals a constant. Your other example doesn't have x squared, so it is not comparable.

The best you could do to factor your equation would be to write it as something like a(x-b)(x-c) = 0; and then b and c will be the two roots found by the quadratic formula.

I see that now, somehow it slipped my mind and I thought I could evade the quadratic formula and factor it somehow. Still thank you very much!

 

[HallsofIvy]

I am confused as to what you were trying to do! You said, originally, "factor" but then you talk about a "linear function". You can't factor a linear function! (Some might say it is "already" factored.) But you can factor a quadradic function so your last post, saying "I thought I could evade the quadratic formula and factor it somehow" makes no sense!

Given any quadratic polynomial, \(\displaystyle ax^2+ bx+ c\), you can solve the equation, \(\displaystyle ax^2+ bx+ c= 0\) for its two solutions, u and v, and then factor it as \(\displaystyle a(x- u)(x- v)\).