MHB Help Needed: Solving for m in an Unfamiliar Practice Problem

[ETBunce]

Not sure exactly what kind of problem I would call this, but I am faced with a strange practice problem in which I must solve for m. The correct answer is apparently 16, but I don't understand how this is true. I am usually left with:
4kx - 8x + 8k = m
and I'm not sure how to simplify further. Here is the problem:

(x - 8)(x - k) = x² - 5kx + m

k and m are constants and the equation is true for all values of x.

Help would be much appreciated. Thanks.

 

[Evgeny.Makarov]

I am usually left with:
4kx - 8x + 8k = m

If for fixed $m$ and $k$ the equality $(4k-8)x=m-8k$ holds for all $x$, what can you conclude?

 

[HallsofIvy]

Not sure exactly what kind of problem I would call this, but I am faced with a strange practice problem in which I must solve for m. The correct answer is apparently 16, but I don't understand how this is true. I am usually left with:
4kx - 8x + 8k = m
and I'm not sure how to simplify further. Here is the problem:

(x - 8)(x - k) = x² - 5kx + m

k and m are constants and the equation is true for all values of x.

Help would be much appreciated. Thanks.

A similar way: Multiplying on the left,
x^2- (8+ k)x+ 8k= x^2- 5kx+ m

Since this to be true for all x, the corresponding coefficients must be the same. So we must have
-(8+ k)= -5k and 8k= m. Solve the first equation for k and put that value into the second equation.

 

[ETBunce]

If for fixed $m$ and $k$ the equality $(4k-8)x=m-8k$ holds for all $x$, what can you conclude?

What should I be trying to conclude from your equation? I don't know where to go from here. Is there a reason you subtracted $8k$ from both sides?

 

[Evgeny.Makarov]

Is there a reason you subtracted $8k$ from both sides?

I just moved all terms containing $x$ to the left-hand side and all other terms to the right-hand side.

What should I be trying to conclude from your equation?

Again, the equation is
\[
(4k-8)x=m-8k
\]
The right-hand side is a fixed number, the left-hand side depends on $x$, and the equation is true for all values of $x$. Suppose the coefficient $4k-8$ of $x$ is nonzero; e.g., $4k-8=3$. Then $x=1$ makes the left-hand side equal 3, $x=2$ makes it equal 6 and so on. Is it possible that all the while the left-hand side stays equal to the fixed value of the right-hand side?

 

[ETBunce]

Is it possible that all the while the left-hand side stays equal to the fixed value of the right-hand side?

I understand HallsofIvy's explanation, so I suppose my question is answered, however, I would like to wrap up this discussion with you.

For the equation $(4k−8)x=m−8k$, it can't be possible that the two sides are the same for all values of $x$, right? Does this mean I simplified incorrectly?

 

[Evgeny.Makarov]

I understand HallsofIvy's explanation, so I suppose my question is answered

I just tried to go deeper into HallsofIvy's claim: "Since this to be true for all x, the corresponding coefficients must be the same". If you take this claim for granted, then $(4k−8)x=m−8k$ implies that $4k-8=0$ since the coefficient of $x$ in the right is $0$. But it's good to understand why the corresponding coefficients must be the same.

For the equation $(4k−8)x=m−8k$, it can't be possible that the two sides are the same for all values of $x$, right?

The two sides can be the same for all $x$, but under one condition. As I explained in post #5, if $4k−8\ne0$, then the equation can't hold for all $x$. It would hold for a single value of $x$ only, namely, $x=(m−8k)/(4k−8)$. Take a couple of minutes to think about the condition I am talking about.

Does this mean I simplified incorrectly?

No, you simplified correctly.

 

[Deveno]

There is another way to do this.

If:

$(x - 8)(x - k) = x^2 - 5kx + m$

for every $x$, then it is certainly true for any particular $x$ we care to choose.

Let's see where choosing $x = 0$ leads us:

$(-8)(-k) = 8k = m$

Ok, now let's see where choosing $x = 1$ leads us:

$(-7)(1 - k) = 1 - 5k + m$

$7k - 7 = m + 1 - 5k$

$12k - 8 = m$.

Now we have two different expressions in $k$ which are both equal to $m$, that is:

$12k - 8 = 8k$.

Can you solve this?

(Note: it really doesn't matter which values we pick for $x$, but the algebra gets "messier" for example, if we pick $x = \pi$ and $x = \sqrt{2}$. Perhaps you are intended to equate the two polynomials:

$(x - 8)(x - k) = x^2 - (k + 8)x + 8k = x^2 - 5kx + m$

and deduce that:

$k + 8 = 5k$
$8k = m$

The first equation tells us that:

$4k = 8$ which when solved for $k$ quickly gives $m$.

But this relies on facts about polynomials you may, or may not know, whereas my approach only requires you to understand that "for all" implies "for any", and thus "for any particular", and we can pick a "particular" that is easy to work with).