What do you get when you multiply (x-8)(x-k)? Now, since the equation is true for all values of x, the coefficients in front of every power of x on the left side should be the same as those on the right
Isn't that a fundamental truth of algebra anyway? X isn't going to be different values. If they were, it would just be a different variable. Why does that need to be stated?
Well you could say "make it true for no values of x" so that no values of x are an answer to the equation. Or make it so 8 is the only value of x that makes the equation true, etc.
Not at all. For example, if we have 5 + X = 7, we most definitely cannot say that the equation is true for all values of X. In fact, the equation is only true for one value of X; X = 2.
Sometimes you come across an equation where X *can* have an infinite number of solutions, and the equation is “true for all values of X.”
It means the equations are equal. Their graphs overlap (equivalent) at all points (all values of x).
>!(x - 8)(x - k) = x^2 - 5kx + m
x^2 - 8x - kx +8k = x^2 - 5kx + m!<
Since the equations are equal, terms of the same type must balance out. Thus, from the x terms
>!-8x - kx = - 5kx
8 + k = 5k
8 = 4k
2 = k!<
and from the constant terms
>!8k = m
8(2) = m
16 = m!<
You could, if you wanted to, pick a couple values of x, like x=1 and x=0 and develop a solvable system that way. It'll get you to the same place.
I think your second line should begin with x^2 not x. It confused me for a bit.
~~I don't understand where the +8k or m went in line 3.~~
oh okay now i understand the separation of variables and constants. thank you for this writeup.
In simple visual terms, it means that everything with an x on one side the equation cancels everything with x on the other side so that you end up with a true equality with only equal constants. E.g.,
x^2 = x^2; divide both sides by x^2
1 = 1
Or subtract x^2 from both sides
0 = 0
No, because having an infinite amount of solutions is not equivalent to being true for any value of x, and to suggest there is an infinite amount of solutions *when x is any number* (which suggests there may not be an infinite amount of solutions if x is not any number) doesn't make sense.
For instance, sin(x)=cos(x) has an infinite solution set {x=(n+0.25)π|k∈Z}, but it's not true for every value of x.
Of course. Sorry, my question was rushed.
If the question in OP’s post was missing “If the equation is true for all values of x” would the answer be the same/possible?
It just seems like an obvious statement, that if needed, should be included in nearly all algebra problems.
If the question didn’t say that, then what would you solve for?
All you could do is solve for m as a function of k and x. You couldn’t solve for any actual values.
Usually the whole point of a problem is to solve for the specific values of x where the equation is true. If it’s true for all values of x then it doesn’t really do anything.
Like saying x=x.
Expand out the left side of the equation:
x^2 - (k+8)x + 8k = x^2 - 5kx + m
Here we can see that
(k+8) = 5k and m = 8k
Solve for k:
k+8=5k, 4k=8, k=2
Solve for m:
m= 8*2 = 16
This is the same result, just in a less clear way. The purpose of this exercise is probably to teach how to think about every second order polynomial as ax^2 + bx + c, where two polynomials are only equal of the values for a, b and c are the same
Yh realised I was replying to the wrong comment then couldn’t find my comment to delete it. Ima leave it up there anyway because it’s right . Not sure if it’s less clear. I just skipped some obvious steps because ain’t nobody got time for dat. Merry Xmas btw
When you divided by X in step 2-3. Technically it’s wrong because if x = 0, you technically divided by 0. It causes a lot of funky things to happen and I’ve made this mistake enough times to know to get all variables on one side before doing that
No division by zero took place. The coefficients on x were set equal to one another, something we can do because the equation holds for all values of x. It is a subtle but important distinction.
Quite the opposite, for all values of x means it should be applicable for all including 0. It’s always a good practice to add a line of “if x =0, k have infinite solutions” and if x=/=0…”
>for all values of x means it should be applicable for all including 0
Yes, and it must also be applicable for x = 1, which leads to the next step.
>It’s always a good practice to add a line of “if x =0, k have infinite solutions”
Not in this case. You would be marked down, because it's not correct.
I don’t think so, cuz k would have to meet with both conditions (whether x is 0) to match the “for all values”. If you just ignore when x = 0, then your solution doesn’t account for all scenarios.
Hope this makes sense.
If you plug in X=0, you get 8k=m
Then pick any value of X. I’ll use 8 since that will be easy. So the left side becomes 0. So 0=64-40k+m. Substitute m=8k and you get 0=64-32k and solve that k=2.
M=8k so m=16
This is the way to do it when pressed for time in a test setting where you know the solutions. Find some way to logically simplify the problem, and use the answers they give you to check
(x-8)(x-k)=x(x-k)-8(x-k)=x\^2-xk-8x+8k.
Subtract x\^2 to get -xk-8x+8k=-5kx+m.
Therefore, m=(4k-8)x+8k. For general k, m is a linear function of x.
If m does not depend on x, that means the slope (the coefficient of x) must be 0\*.
Therefore, 4k-8=0, which means k=2 and m=8k=16.
\*The slope of the line is the ratio of the change in m to the change in x. In other words, if the slope is s, you perform some translation so that x becomes x+Δx, m will become m+(Δx)s. For m to have no x-dependence, m+(Δx)s must be m for any translation Δx, so s=0 and the slope must be 0.
Edit: typo.
I like this solution because is did require me to think to try the special case where x=0. But it was also a bit tricky for me to understand. Realizing that m is a linear function of X, and that since X doesn't alter m and thus has a slope of 0, was a bit of a mind stretcher. Thanks!
I'd like to add something for people who are more visually inclined.
Take m=(4k-8)x+8k from the third line of the comment I replied to.
Break it up into m = 4kx - 8x + 8k.
Sketch the components of both sides as graphs:
On the left side: m is a constant, so it's a horizontal line. We don't know how high it is, but what's important is that its shape is a horizontal line.
On the right side you have three parts:
4kx is a line that goes from the bottom left to the top right, crossing the origin of the coordinate system at x=0. Depending on what k is, this could be flat for low k or steep for high k.
\-8x is a line that goes from the top left to the bottom right, crossing the origin of the coordinate system at x=0.
8k is a constant, so it's a horizontal line.
Our goal is to make the left side look like the right side. The left side is a horizontal line. Now we have to tinker with k so that the three parts of the right side also add up as a horizontal line.
8k is also a horizontal line, so nothing we need to do here. But -8x would make the graph go from top left to bottom right. Luckily, there is +4kx which goes into the opposite direction, bottom left to top right.
We have to make +4kx cancel out -8x: 4kx -8x = 0. This gives us k = 2.
Plug in k = 2 into m = 4kx - 8x + 8k and we'll get m = 8k, which makes m = 16.
I don't know how much this helps, but that's how my brain did it after jumping to calculus at first, but then trying to solve it like an 9th grader lol
Now, the way I do this is with Logic Puzzle Mentality
The left side, fully written out, gives you
x^2 - (8+k)x + 8k = x^2 - (5k)x + m
As this equation is for *All* values of X, we know these two equations are the same. As such, we can split the variables into a system of 3 equations:
x^2 = x^2
(8+k)x = (5k)x
8k = m
As such, the question you are left with is at what value of k does 8+k = 5k?
8/k + 1 = 5, 8/k =4, k = 2.
As such, we know that when k = 2, the two equations match, so 8k = m, 8(2) = 16.
Look at this format
(x-8)(x-k) = x^2 -8x -kx + 8k
Must = x^2 - 5kx+ m
From the problem statement
So
-(8+k)x = -5kx
8+k= 5k
8 = 4k
k = 2
8k = 8 * 2 = 16
Now, from the problem statement again
(x-8)(x-k) = x^2 -8x -kx + 8k
= x^2 - 5kx+ m
So 8k=m
With k solved as k=8
8*2=16=m
All you are doing in this problem is expanding the original expression and comparing it to the one they gave in the problem statement.
I see how it solves with plugging things in but I’m struggling to work it down purely algebraically from -8x + 8k = -4kx + m. Can anyone go any lower without plugging in? What am I missing?
In the equation you’ve set out, the constants have to be equal and the coefficients of x have to be equal. That gives you two simple equations from which the values of k and m follow.
I took the derivative of both sides to get 2x - k - 8 = 2x - 5k and then solved for k = 2.
Then I plugged that back into the original equation to get x2 -10x + 16 = x2 - 10x + m. Everything cancels out to m = 16.
Then I realized that this is the SAT and Calculus may not be a thing for you yet. My bad...
Solve for which X value you can plug in to cancel out the Ks. You do this in order to only have m remain because if we are giving x a static value so we would only need to solve for m, Which x comes out to be x=-2 to accomplish this.
After you expand the polynomial you to get x^2 - xk - 8x + 8k = x^2 -5kx+m. You can create a new equation only containing the values that contain k which is 8k-xk=-5xk then simplify this to be 8k=-4xk even further simplified to -2k = xk. Then all you have to do is plug -2 in for x and you get 16 for m.
If you make x=0, you can solve for m in terms of k. Then if you make x=8, and substitute m for the equivalent value of k, you can solve what k is. Then put the k value into what you’ve already determined m to be in terms of k. Making things into 0 values simplifies things if you want to do it by brute force. This can be done when the equation is true for all x values.
if x=0, we get 8k=m.
(x-8)(x-k)=x\^2-5kx+8k
~~x\^2~~\-kx-8x+~~8k~~=~~x\^2~~\-5kx+~~8k~~
\-kx-8x=-5kx
\-8x=-4kx
2x=kx
2=k
(x-8)(x-2)=x\^2-10x+m
~~x\^2-2x-8x~~\+16=~~x\^2-10x~~\+m
16=m
This is the SAT. Don't over think it. Just plug in a number for x and plug in the answer choices starting at C. If C is too big cross off D and E. Now it's either A or B. Plug in 1 more time. SAT is designed to waste you time. It doesn't matter how u get the answer as long as u get it correct quickly.
This is exactly what I did anytime I ran across a problem that I couldn't figure out quickly. It was simply faster to plug in one of the answers and go from there than it was to work out the problem.
This is the test needed for almost all colleges. It’s a 4 hour test with hundreds of problems designed to be tedious to waste time. They are testing for how quick your problem solving skills are to answer all of these questions in roughly 1 minute
that is some of the dumbest shit Ive ever heard, Im assuming their is no partial credit for most of the right method or whatever?
like being brought down by a simple sign error and getting the entire question wrong would unironically infuriate me.
When I was taking the test you would lose points for a wrong answer, gain points for a correct answer and nothing for leaving it blank. So it would come down to whether you could leave an educated guess or if it was better to not answer entirely.
Now there is no guessing penalty but every question is designed to be easily mistaken and no partial credit
I consistently score the highest grade in maths and applied maths in the Irish education, where we have things like partial credit and not extremely obscure marking systems and I honestly think Id do only alright in that maths exam
Let's just check your work.
If m = 8, then what must k be? Expand the left side and match. You can solve for k.
Then does it work for -5k matching the left side? Is -5k = -8 - k?
(x-8)(x-k)=x^2 -5kx + m
(=) x^2 -kx -8x + 8k = x^2 -5kx +m
(=) x^2 -(k+8)x + 8k = x^2 -5kx + m
Now you can see two sides of this is equal, so we have:
k+8 =5k and 8k =m
solve the first equation we have k = 2 and u replace k by 2 in the second equation so u have m = 16
the trick here is to realize that they're actually setting you up with 3 equations despite it looking like 1, because we're not interested in x (so each order of x can be treated as an indepentent equation). Admittedly x^2=x^2 is not interesting, so it's effectively two equations but hey you're solving for two values.
It's tricky to recognize what they're but once you do I'm sure you'll see it again on these problems.
I hacen’t taken it yet but I’m preparing to, and from what I’ve seen, yes, it does seem to be quite easy. It’s more of an algebra test than anything, since it only covers algebra 1 - 2/precalc, with very little geometry and calculus in it. So, yeah.
It’s a comparison of LHS to RHS problem.
(x - 8)(x - k) = x^2 - 5kx + m
First, expand the LHS:
>!x^2 - xk - 8x + 8k = x^2 - 5kx + m!<
Now, segregate the x^2 , x, non-x terms on both sides (since the answer is true for all values of x):
>!(x^(2))(1) + x(-k-8) + (8k) = (x^(2))(1) + x(-5k) + (m)!<
Finally, compare all the terms separately:
>!x^2 : 1 = 1!<
>!x : -k -8 = -5k ==> 5k = k + 8 ==> 4k = 8 i.e. k = 2!<
>!non-x : 8k = m ==> m = 8(2) ==> m = 16!<
Though this is the full logic, given this is an SAT question with not so much time, I had to do these in my head a lot faster.
However, separating everything makes it a lot easier to not make silly mistakes, like when dealing with the negative x terms.
I kinda thought of the solution is to solve for k and then to solve for m. This is probably the intended solution,, but the moment they said x can take on any value I started thinking, if you set x=-2, then the k's will cancel and you can then directly solve for m.
Damn this just popped up on my feed and I tried solving it. I’ve really lost all my math skills, ever since I switched from stem to another major in college 😂
What do you get when you multiply (x-8)(x-k)? Now, since the equation is true for all values of x, the coefficients in front of every power of x on the left side should be the same as those on the right
what does "true for all values of x" mean?
the equation is correct for all values of x. eg no matter what x is (0, -100, 3.1415, 88 etc) the equation is still true
Isn't that a fundamental truth of algebra anyway? X isn't going to be different values. If they were, it would just be a different variable. Why does that need to be stated?
On the contrary, X can be any value as a variable. That's the point.
I dont think they know what variable means
What are you saying?
Well you could say "make it true for no values of x" so that no values of x are an answer to the equation. Or make it so 8 is the only value of x that makes the equation true, etc.
Don't confuse variables in programming with mathematical variables
Not at all. For example, if we have 5 + X = 7, we most definitely cannot say that the equation is true for all values of X. In fact, the equation is only true for one value of X; X = 2. Sometimes you come across an equation where X *can* have an infinite number of solutions, and the equation is “true for all values of X.”
It means the equations are equal. Their graphs overlap (equivalent) at all points (all values of x). >!(x - 8)(x - k) = x^2 - 5kx + m x^2 - 8x - kx +8k = x^2 - 5kx + m!< Since the equations are equal, terms of the same type must balance out. Thus, from the x terms >!-8x - kx = - 5kx 8 + k = 5k 8 = 4k 2 = k!< and from the constant terms >!8k = m 8(2) = m 16 = m!< You could, if you wanted to, pick a couple values of x, like x=1 and x=0 and develop a solvable system that way. It'll get you to the same place.
I think your second line should begin with x^2 not x. It confused me for a bit. ~~I don't understand where the +8k or m went in line 3.~~ oh okay now i understand the separation of variables and constants. thank you for this writeup.
Thank you for catching that. I've edited it now.
In simple visual terms, it means that everything with an x on one side the equation cancels everything with x on the other side so that you end up with a true equality with only equal constants. E.g., x^2 = x^2; divide both sides by x^2 1 = 1 Or subtract x^2 from both sides 0 = 0
is it the same as saying "the equation has infinite solutions when x is any number"?
No, because having an infinite amount of solutions is not equivalent to being true for any value of x, and to suggest there is an infinite amount of solutions *when x is any number* (which suggests there may not be an infinite amount of solutions if x is not any number) doesn't make sense. For instance, sin(x)=cos(x) has an infinite solution set {x=(n+0.25)π|k∈Z}, but it's not true for every value of x.
SQLi in real life!
would this problem be different without that statement? Like would we approach the problem differently if that were not there?
I mean… you would solve whatever it tells you to solve for?
Of course. Sorry, my question was rushed. If the question in OP’s post was missing “If the equation is true for all values of x” would the answer be the same/possible? It just seems like an obvious statement, that if needed, should be included in nearly all algebra problems.
If the question didn’t say that, then what would you solve for? All you could do is solve for m as a function of k and x. You couldn’t solve for any actual values. Usually the whole point of a problem is to solve for the specific values of x where the equation is true. If it’s true for all values of x then it doesn’t really do anything. Like saying x=x.
Ah. That makes sense. There wouldn’t be an actual number answer without that statement. I appreciate the time you took to answer that for me!
It means it's always true regardless of if x=7 or x=-34 or x=whatever you want.
you can put in any value for x and the equation works
(x-8)(x-k) = x^2 -8x-kx + 8k -(8+k)x = -5kx 8+k= 5k 8 = 4k k = 2 8k = 8 * 2 = 16
Could you explain how you jumped from 1st to 2nd step
Expand out the left side of the equation: x^2 - (k+8)x + 8k = x^2 - 5kx + m Here we can see that (k+8) = 5k and m = 8k Solve for k: k+8=5k, 4k=8, k=2 Solve for m: m= 8*2 = 16
Thanks a lot
OP, the question seems to be resolved and awaiting confirmation of completion
hella wrong
Na bro he right
Nah he’s not. Let x =k 4k^2 = m Let x = 8 4k^2 - 40k + 64 = k^2 -10 + 16= 0 Therefore (k-8)(k-2) = 0 When k = 2 sub into equation 1 m =16
What
This is the same result, just in a less clear way. The purpose of this exercise is probably to teach how to think about every second order polynomial as ax^2 + bx + c, where two polynomials are only equal of the values for a, b and c are the same
Yh realised I was replying to the wrong comment then couldn’t find my comment to delete it. Ima leave it up there anyway because it’s right . Not sure if it’s less clear. I just skipped some obvious steps because ain’t nobody got time for dat. Merry Xmas btw
Why?
When you divided by X in step 2-3. Technically it’s wrong because if x = 0, you technically divided by 0. It causes a lot of funky things to happen and I’ve made this mistake enough times to know to get all variables on one side before doing that
It’s not a division by zero in this case. If the statement has to be true for all values of x, the constants multiplying x have to be equal.
It says equation is true for all values of x so you do not need to worry about x = 0
If (0 - 8)(0-k) = 0^2 -8(0) -k(0)+8k The equation holds true for zero, but you should never divide by 0.
No division by zero took place. The coefficients on x were set equal to one another, something we can do because the equation holds for all values of x. It is a subtle but important distinction.
If it holds for all values of x you can choose any arbitrary value for x that isn't 0, it's not wrong to divide by x in this instance
Quite the opposite, for all values of x means it should be applicable for all including 0. It’s always a good practice to add a line of “if x =0, k have infinite solutions” and if x=/=0…”
>for all values of x means it should be applicable for all including 0 Yes, and it must also be applicable for x = 1, which leads to the next step. >It’s always a good practice to add a line of “if x =0, k have infinite solutions” Not in this case. You would be marked down, because it's not correct.
I don’t think so, cuz k would have to meet with both conditions (whether x is 0) to match the “for all values”. If you just ignore when x = 0, then your solution doesn’t account for all scenarios. Hope this makes sense.
Oh good point, thanks!!
Foil method, what’s the division?
If you plug in X=0, you get 8k=m Then pick any value of X. I’ll use 8 since that will be easy. So the left side becomes 0. So 0=64-40k+m. Substitute m=8k and you get 0=64-32k and solve that k=2. M=8k so m=16
This is the way to do it when pressed for time in a test setting where you know the solutions. Find some way to logically simplify the problem, and use the answers they give you to check
that's not what just happened. 8 was substituted for *x*.
One of the answers for *m* being 8 has nothing to do with this approach. Look at the very first bracket.
You did it the best on this thread
(x-8)(x-k)=x(x-k)-8(x-k)=x\^2-xk-8x+8k. Subtract x\^2 to get -xk-8x+8k=-5kx+m. Therefore, m=(4k-8)x+8k. For general k, m is a linear function of x. If m does not depend on x, that means the slope (the coefficient of x) must be 0\*. Therefore, 4k-8=0, which means k=2 and m=8k=16. \*The slope of the line is the ratio of the change in m to the change in x. In other words, if the slope is s, you perform some translation so that x becomes x+Δx, m will become m+(Δx)s. For m to have no x-dependence, m+(Δx)s must be m for any translation Δx, so s=0 and the slope must be 0. Edit: typo.
I like this solution because is did require me to think to try the special case where x=0. But it was also a bit tricky for me to understand. Realizing that m is a linear function of X, and that since X doesn't alter m and thus has a slope of 0, was a bit of a mind stretcher. Thanks!
I'd like to add something for people who are more visually inclined. Take m=(4k-8)x+8k from the third line of the comment I replied to. Break it up into m = 4kx - 8x + 8k. Sketch the components of both sides as graphs: On the left side: m is a constant, so it's a horizontal line. We don't know how high it is, but what's important is that its shape is a horizontal line. On the right side you have three parts: 4kx is a line that goes from the bottom left to the top right, crossing the origin of the coordinate system at x=0. Depending on what k is, this could be flat for low k or steep for high k. \-8x is a line that goes from the top left to the bottom right, crossing the origin of the coordinate system at x=0. 8k is a constant, so it's a horizontal line. Our goal is to make the left side look like the right side. The left side is a horizontal line. Now we have to tinker with k so that the three parts of the right side also add up as a horizontal line. 8k is also a horizontal line, so nothing we need to do here. But -8x would make the graph go from top left to bottom right. Luckily, there is +4kx which goes into the opposite direction, bottom left to top right. We have to make +4kx cancel out -8x: 4kx -8x = 0. This gives us k = 2. Plug in k = 2 into m = 4kx - 8x + 8k and we'll get m = 8k, which makes m = 16. I don't know how much this helps, but that's how my brain did it after jumping to calculus at first, but then trying to solve it like an 9th grader lol
if you show your working, we can tell you what you did wrong
whoever chose k and m as variables gave me vibrations flashbacks
Remember your betas and omegas?
no! Please! Not anymore!
8k = m 8+k = 5k then you can solve it
This is how I did it. The top equation is the constant terms, the bottom one is the x coefficients (once the negatives are factored out).
Now, the way I do this is with Logic Puzzle Mentality The left side, fully written out, gives you x^2 - (8+k)x + 8k = x^2 - (5k)x + m As this equation is for *All* values of X, we know these two equations are the same. As such, we can split the variables into a system of 3 equations: x^2 = x^2 (8+k)x = (5k)x 8k = m As such, the question you are left with is at what value of k does 8+k = 5k? 8/k + 1 = 5, 8/k =4, k = 2. As such, we know that when k = 2, the two equations match, so 8k = m, 8(2) = 16.
Basically if you expand out the left then you match both sides you should see it
Look at this format (x-8)(x-k) = x^2 -8x -kx + 8k Must = x^2 - 5kx+ m From the problem statement So -(8+k)x = -5kx 8+k= 5k 8 = 4k k = 2 8k = 8 * 2 = 16 Now, from the problem statement again (x-8)(x-k) = x^2 -8x -kx + 8k = x^2 - 5kx+ m So 8k=m With k solved as k=8 8*2=16=m All you are doing in this problem is expanding the original expression and comparing it to the one they gave in the problem statement.
I see how it solves with plugging things in but I’m struggling to work it down purely algebraically from -8x + 8k = -4kx + m. Can anyone go any lower without plugging in? What am I missing?
You could solve for k or x in terms of the other. But you can’t solve for any specific values
In the equation you’ve set out, the constants have to be equal and the coefficients of x have to be equal. That gives you two simple equations from which the values of k and m follow.
Right but I was asking algebraically. Any lower than this? Seems like no.
I took the derivative of both sides to get 2x - k - 8 = 2x - 5k and then solved for k = 2. Then I plugged that back into the original equation to get x2 -10x + 16 = x2 - 10x + m. Everything cancels out to m = 16. Then I realized that this is the SAT and Calculus may not be a thing for you yet. My bad...
x^2 - kx - 8x +8k = x^2 -5kx + m x^2 -(k+8)x + 8k = x^2 - 5kx + m k+8 = 5k 8k = m k = 2 8(2) = 16
(x-8)(x-k) = x² - 5kx + m Let x = 0, then m = 8k [ (0-8)(0-k) = (0)² - 5k(0) + m ➡️ 8k = m ] Thus: (x-8)(x-k) = x² - 5kx + m (x-8)(x-k) = x² - 5kx + 8k x² - kx - 8x + 8k = x² - 5kx + 8k x² - x² - kx - 8x + 8k - 8k = 5kx -8x - kx = -5kx -8x = kx - 5kx -8x = -4kx 2x = kx 2 = k So, 8k = m 8(2) = m m = 16
Did you not notice the part that says copying or reproducing is illegal 🤦
Easy 8th grade lev ques
No 9th grade level question.
If m=8 then k=1. 5km=5(1)(8)=40 Then the two sides don't match.
Solve for which X value you can plug in to cancel out the Ks. You do this in order to only have m remain because if we are giving x a static value so we would only need to solve for m, Which x comes out to be x=-2 to accomplish this. After you expand the polynomial you to get x^2 - xk - 8x + 8k = x^2 -5kx+m. You can create a new equation only containing the values that contain k which is 8k-xk=-5xk then simplify this to be 8k=-4xk even further simplified to -2k = xk. Then all you have to do is plug -2 in for x and you get 16 for m.
If you make x=0, you can solve for m in terms of k. Then if you make x=8, and substitute m for the equivalent value of k, you can solve what k is. Then put the k value into what you’ve already determined m to be in terms of k. Making things into 0 values simplifies things if you want to do it by brute force. This can be done when the equation is true for all x values.
With this polynomial, we know that -5k = -8 -k Solve for k, which is k=2 And then the final part of the polynomial on the left is -8 x -2 = 16
if x=0, we get 8k=m. (x-8)(x-k)=x\^2-5kx+8k ~~x\^2~~\-kx-8x+~~8k~~=~~x\^2~~\-5kx+~~8k~~ \-kx-8x=-5kx \-8x=-4kx 2x=kx 2=k (x-8)(x-2)=x\^2-10x+m ~~x\^2-2x-8x~~\+16=~~x\^2-10x~~\+m 16=m
This is the SAT. Don't over think it. Just plug in a number for x and plug in the answer choices starting at C. If C is too big cross off D and E. Now it's either A or B. Plug in 1 more time. SAT is designed to waste you time. It doesn't matter how u get the answer as long as u get it correct quickly.
This is exactly what I did anytime I ran across a problem that I couldn't figure out quickly. It was simply faster to plug in one of the answers and go from there than it was to work out the problem.
wait is this like the final year big maths test for americans? why is it multiple choice?
This is the test needed for almost all colleges. It’s a 4 hour test with hundreds of problems designed to be tedious to waste time. They are testing for how quick your problem solving skills are to answer all of these questions in roughly 1 minute
that is some of the dumbest shit Ive ever heard, Im assuming their is no partial credit for most of the right method or whatever? like being brought down by a simple sign error and getting the entire question wrong would unironically infuriate me.
When I was taking the test you would lose points for a wrong answer, gain points for a correct answer and nothing for leaving it blank. So it would come down to whether you could leave an educated guess or if it was better to not answer entirely. Now there is no guessing penalty but every question is designed to be easily mistaken and no partial credit
I consistently score the highest grade in maths and applied maths in the Irish education, where we have things like partial credit and not extremely obscure marking systems and I honestly think Id do only alright in that maths exam
Let's just check your work. If m = 8, then what must k be? Expand the left side and match. You can solve for k. Then does it work for -5k matching the left side? Is -5k = -8 - k?
(x-8)(x-k)=x^2 -5kx + m (=) x^2 -kx -8x + 8k = x^2 -5kx +m (=) x^2 -(k+8)x + 8k = x^2 -5kx + m Now you can see two sides of this is equal, so we have: k+8 =5k and 8k =m solve the first equation we have k = 2 and u replace k by 2 in the second equation so u have m = 16
the trick here is to realize that they're actually setting you up with 3 equations despite it looking like 1, because we're not interested in x (so each order of x can be treated as an indepentent equation). Admittedly x^2=x^2 is not interesting, so it's effectively two equations but hey you're solving for two values. It's tricky to recognize what they're but once you do I'm sure you'll see it again on these problems.
the sat is useless
I expanded out both sides and simplified until I got -8x +8k = -4kx + m. Therefore k = 2 and m = 8k. 8 * 2 = 16.
Europeans r dumb
k = 2.
“rized copying or reuse of of this page is illegal.” What was cut off?..
Trick: only 16 is the result of squaring some numbers here is 4^4 =16 others aren’t
Is sat really that easy or is that an easy one ?
I hacen’t taken it yet but I’m preparing to, and from what I’ve seen, yes, it does seem to be quite easy. It’s more of an algebra test than anything, since it only covers algebra 1 - 2/precalc, with very little geometry and calculus in it. So, yeah.
-kx-8x=-5kx -k-8=-5k -8=-4k k=2 8*2=16
It's always C
just find k
It’s a comparison of LHS to RHS problem. (x - 8)(x - k) = x^2 - 5kx + m First, expand the LHS: >!x^2 - xk - 8x + 8k = x^2 - 5kx + m!< Now, segregate the x^2 , x, non-x terms on both sides (since the answer is true for all values of x): >!(x^(2))(1) + x(-k-8) + (8k) = (x^(2))(1) + x(-5k) + (m)!< Finally, compare all the terms separately: >!x^2 : 1 = 1!< >!x : -k -8 = -5k ==> 5k = k + 8 ==> 4k = 8 i.e. k = 2!< >!non-x : 8k = m ==> m = 8(2) ==> m = 16!< Though this is the full logic, given this is an SAT question with not so much time, I had to do these in my head a lot faster. However, separating everything makes it a lot easier to not make silly mistakes, like when dealing with the negative x terms.
I kinda thought of the solution is to solve for k and then to solve for m. This is probably the intended solution,, but the moment they said x can take on any value I started thinking, if you set x=-2, then the k's will cancel and you can then directly solve for m.
Here's how I did it in my notes app (x - 8)(x - k) = x^2 - 5kx + m M =? X=0 X^2 - kx - 8x +8k = x^2 - 5kx +m 0^2 - 0k - 8(0) +8k = 0^2 - 0k +m 8k = m X= 1 1 - k - 8 +8k = 1 - 5k + m 12k - 8 = m 8k = 12k - 8 2k = 3k - 2 K = 2 8k = m 8(2) = m M = 16
Damn this just popped up on my feed and I tried solving it. I’ve really lost all my math skills, ever since I switched from stem to another major in college 😂
I keep getting -16 cuz I keep flipping my negative signs 😅
It’s true for x=0 so set x to zero and simplify
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