Question

No idea how to do this! (x-8)(x-k) = x^2 - 5kx +m
In the equation above, k and m are constants. If the equation is true for all values of x, what is the value of m?
((I'm an International student and I've never been asked a question like this..))

Answer 1

\[(x-8)(x-k) = x ^{2} - 5kx + m\]

Answer 2

true for all \(x\)? then try \(x=0\)

Answer 3

How does saying "true for all x" denote that x = 0? My apologies, I just haven't done this is my own school before!

Answer 4

it says "true for all \(x\)" apparently, so what i am saying is that if it is true no matter what \(x\) is, then it must be true if \(x=8\)

Answer 5

if you replace \(x\) by \(8\) you get
\[64-40k+m=0\] as one equation or
\[40k-m=64\] then pick some other value for \(x\) get another equation with in \(k\) and \(m\) and solve

Answer 6

it doesn't really matter what value of \(x\) you pick, pick any two and solve

Answer 7

I figured out what m equals

Answer 8

I think it is wise to let x equal 8 then create the first equation. Afterwards, let x = k, then create the second equation.

Answer 9

@Hero @satellite73 Thank you both; just another question: So why pick 8? Could you pick let's say 6, or 4? And what "type" of Math is this? (ie: Advanced Algebra, Geometry, Calculus etc..)? Thanks

Answer 10

You use x = 8 and x = k because you can simplify the equations better that way

Answer 11

For example on the left side:

When you use x = 8, the left side reduces to 0:
When you use x = k, the left side also reduces to 0:

This makes solving for m more convenient.

Answer 12

@Dj2k12

Answer 13

@Hero Ah, okay. I'm going to try and work this out, and I'll publish what I end up with here. Thanks for your help!

Answer 14

\[(x-8)(x-k)=x ^{2}-5kx+m. \]
So from here, I plug "8" in on the left side.
\[(8-8)(8-k) = 0\]
Now I have to plug 8 into "x" on the other side too, so:
\[x ^{2} - 5k(8) +m = 0\] Or.. \[64-40k+m=0\]
So how do I get rid of k here? Am I right so far?

Answer 15

@Hero

Answer 16

It looks right. Now do it again except replace x with k

Answer 17

So if I replace k with x, and x = 8, then do I just replace k with 8? Because when I do that I get -256. @Hero

Answer 18

Just do the same thing you did the first time but replace x with k ONLY

Answer 19

you don't replace k with 8

Answer 20

I'm not entirely getting it. I don't have anything to represent k so how can I replace x with it? @Hero. Sorry, as I said, we don't cover this type of Math in Ireland lol

Answer 21

What is so difficult about replacing x with k everywhere you see x?

Answer 22

pretend that k is a number and replace it with x the same way you did 8 the first time.

Answer 23

Erm, plugging in isn't difficult, it's understand why and when that's a challenge. Regardless, I'll give it a shot:
\[x ^{2}-5kx+m\]
Now I'm replacing x with k.
\[k ^{2}-5k(k) + m = 0\]
\[k^2 - 5k^2 +m= 0\]
\[-4k^2+m=0\]
.. Now what do I do?

Answer 24

@Hero

Answer 25

add \(4k^2\) to both sides

Answer 26

\[m= +4k^2?\]

Answer 27

Yes, now set m = m and solve for k

Answer 28

After you find k, plug the value you find back in to find m. (Systems of Equations)

Answer 29

Okay, so firstly, I'm plugging my new value for M into the equation to solve for K:
\[k^2 - 5k^2 + 4k^2 = 0\]
And now I have 0=0.

Answer 30

You're ignoring the first equation you created when you had \(m = 40k - 64\)

Answer 31

I stated to set m = m, then after doing that, solve for k.

Answer 32

But that's because I found out that m = 4k^2?

Answer 33

Oh I see what you mean, hold up

Answer 34

smh

Answer 35

so \[4k^2 = 40k-64\]
Divide across by k
\[4k= 40-64\]
4k= 24
and thus, k = 6.
Right so far? @Hero

Answer 36

Never ever divide by the variable you are solving for

Answer 37

@Hero Sorry, I'm just not getting this question. Thanks anyway.

Answer 38

We're almost done. Why are you going to give up now?

Answer 39

Clearly we have a quadratic equation left to solve.

Answer 40

Because I'm utterly lost. But I guess I can keep trying. Was I right with k= 4?

Answer 41

Oops, I mean 6. ^^

Answer 42

You're not lost. You made it this far without asking for much help.

Answer 43

I'm just confused because first m= 40k-64, and then it equals \[4k^2\] and I finally get k = 6 and I'm not sure where to go from here, and why I am still ignoring the first part of the equation.

Nope, apparently I did not.. but I'm guess that I should bring the 4k^2 over the equals and use the -b formula?

Answer 44

My first instinct would be to divide across by x so that I could have it on the left side and allow it to equal to an integer on the right; but you said to "never divide by the variable you're trying to solve" so my next guess would be to find the factors of 64 and factorize the equation, or else use the -B formula. Sorry, but as you can see, Math isn't one of my strengths.

Answer 45

Okay, the correct quadratic equation should be

\[4k^2 - 40k + 64 = 0\]

Answer 46

Writing it this way, there's no way you can simply divide both sides by k

Answer 47

Dividing by k eliminates one of the solution values. You do not want to eliminate solution values.

Answer 48

Of course I have, but we tend to use the -B formula to sort out all Quadratic equations by using the formula \[ax^2 + bx+ c\] and plugging the numbers into the minus B formula. Regardless, I could do the same here by allowing
a = 4
b= 40 and
c = 64.

Or else I could say
(4k+16)(k-4) or something like that.. I'm not entirely certain @Hero

Answer 49

Oh wait, I factorized wrongly. So -B would be my last resort.

Answer 50

That -b formula you're referring to is called the Quadratic Formula by the way and b = -40

Answer 51

But anyway, you do not have to use the Quadratic Formula if the quadratic equation is factorable.

Answer 52

Listen, thanks for your help but I really am lost in this question. I'm going to try and perfect the Medium and Easy questions and perhaps skip the questions like these on the SAT. But again, thanks for your help up until this point but I'm just not getting this. @Hero

Answer 53

\[4k^2 - 40k + 64 = 0
\\4(k^2 - 10k + 16) = 0
\\k^2 - 10k + 16 = 0
\\k^2 - 8k - 2k + 16 = 0
\\k(k - 8) -2(k - 8) = 0
\\(k - 8)(k - 2) = 0
\\k = 2
\\ k = 8
\]

Answer 54

Using k = 2, you get m = 16

Answer 55

@Hero. Thank you; I'll write this down and I'll try to maybe work it out on paper until it makes sense. Again, thanks for your help and patience.

Answer 56

Well unfortunately it is not as simple as that. You'll probably get more confused before you understand it. I think you should take satellite's advice and try different values of x. It's kinda risky though.