<p>I have a no idea what a constant is and I've seen it in quite a few SAT questions. What rules apply when the question tells you that a variable is a constant? Here is a sample question that I have no clue how to do, if anyone can solve this and explain Id be very grateful, thanks.</p>
<p>(X-8)(x-k)=x^2 - 5kx + m</p>
<p>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?</p>
<p>(x-8)(x-k) = x^2-5kx+m
x^2 - kx - 8x + 8k = x^2 - 5kx + m
x^2 - x(k + 8) + 8k = x^2 - 5kx + m </p>
<p>for the equation to be true
-x(k+8) = -5kx and 8k = m
k+8 = -5k and m = 8k
6k =-8
k = -4/3 </p>
<p>m = 8(-4/3) = -32/3 </p>
<p>edit: </p>
<p>How I did it? I multiplied out (x-8) and (x-k) then realized that they are both quadratic and if they are both quadratic and equal, everything should be equal. So I set up the first equation into a conventional quadratic and set equal the two parts of it and solved. k solved easier and then you plug and chug.</p>
<p>Ok thank you! But could you explain how you got from -x(k+8)+8k= -5kx+m to
-x(k+8)=-5kx and 8k=m?
Is that an algebra rule that you can split an equation into 2? </p>
<p>small correction:
-x(k+8) = -5kx
k+8 = 5k, not -5k
–> k = 2 and m = 16.</p>
<p>Sethrogan, a constant is just an unvarying value. For example, pi is a constant (3.14159…) and the number of inches in a foot is a constant (12).</p>
<p>In this problem, k and m are constants: 2 and 16. Contrast these with x, which is not a constant; it can be changed to any value.</p>
<p>Constant: a number or variable that has fixed value - in this case, treat k and m as if they were numbers. In general, if something is constant, it doesn’t change.</p>
<p>Equation holds for <em>all</em> values of x. If x^2 - (k+8)x + 8k = x^2 - 5kx + m, then the coefficients of x^2, x, and 1 must necessarily be equal, so -(k+8) = -5k and 8k = m (same as what livelaugh7 said).</p>
<p>simplify line 2 from livelaugh and you get
4xk - 8x + 8k - m = 0 or
4xk - 8x = m - 8k</p>
<p>This is true “for all values of x”</p>
<p>at x = 0: 0 = m - 8k or m=8k
at x = 2: k = 2 (thus m=16)
at x = -3: 4(-3)(2) - 8(-3) = 16 - 8(2)
-24 + 24 = 0</p>
<p>So, m=16 and k=2</p>
<p>I’ll double check when I can . . . . key is that this is true for ALL values of x and k & m do not change. You use this in trigonometry.</p>
<p>I really don’t understand this. I’m sorry guys I really just don’t think my algebra is good enough for me to comprehend any of this. Does anyone have an alternate way of doing this? Or can someone explain in a really simple way and explain what rules they’re using when they do a step? Thanks for the help so far guys.</p>
<p>When the equation says the equation is true for all values of X, does that mean that you can plug in ANY number for x? If so how do you know what the right number to plug into x is?</p>
<p>Can anyone please answer my second question? Why can you split -x(8+k)+8k=-5kx+m into -x(8+k)=-5kx
And 8k=m??? Is there some rule I’m missing here?</p>
<p>Yes - you can plug in ANY number for x and it will work. There is no “correct” value for x. Take the equation “x = x” for example. Is this true for any value of x? Let’s try some values.</p>
<p>x = x ???
5 = 5 yep
10 = 10 yes…
32 billion = 32 billion</p>
<p>It appears so! ANY value of x will work in this equation. There is no “right number” for x - ANY value works. It DOES NOT MATTER what x is.</p>
<p>In your problem, you are trying to navigate the original equation
(x-8)(x-k)=x^2 - 5kx + m
into a form where the left side is the same as the right side; if they are the same, then ANY value of x will work, just like x = x.</p>
<p>To do this, we follow the steps in the previous posts to get to the point where you got stuck earlier:
x^2 - x(k + 8) + 8k = x^2 - 5kx + m</p>
<p>Remember that the left side should be the same as the right side. Ask yourself: what values do I stick into k and m so that the two sides are the same? Notice how each side has terms of x^2, x, and 1. The amounts of these on the two sides must be equal for the two sides to be the same. How many of these are on each side?</p>
<p>x^2
It looks like we have one x^2 on each side. 1 = 1. The two sides are equal - we like that.</p>
<p>x
We have -x(k+8) on one side and -5kx on the other, so on one side there are -(k+8) “x’s” and on the other side there are -5k “x’s”. Remember, the number of x’s must be equal, so:
-(k+8) = -5k --> k + 8 = 5k --> 8 = 4k --> k = 2</p>
<p>1
We have 8k “1s” on the left and m “1s” on the right. These must be equal.
8k = m</p>
<p>Since we already know that k = 2 from examining the x’s we can just plug it into the 1s: m = 8(2) = 16.</p>
<p>As you said: if “you can plug in ANY number for x” that means there is NO CORRECT VALUE for x AND NO INCORRECT VALUE. ALL values and any value is true. Picture a graph, anywhere there is a value of x, there is a corresponding value for y (hence a point).</p>
<p>I solved it in post #5 by substituting in different values for x and getting values of (“solving for”) k & m.</p>
<p>FYI, if you take each question on Math SAT and interpret it exactly as it says, it will do you good. If it says “equation is true for all values of x”, it means exactly what it means. Therefore it is true if x = 3, or x = 0, or x = 10^9.</p>
<p>The best value to plug in for x (if it holds for all x) is any x that makes the problem easier for you to do.</p>
<p>Ohhh wow thank you so much guys!
I had no clue you could separate x^2’s and x’s values and set them equal when dealing with trinomials. Thank you, especially NHTLDJ, you made me understanding what was completely stumping me. Thanks again everyone!</p>
<p>Ooops. My math was a bit off. Excuse me.</p>