<p>if k and h are constants and x^2+kx+7 is equivalent to (x+1)(x+h), what is the value of k??</p>
<p>a 0
b 1
c 7
d 8
e It cannot be determined from the information given.</p>
<p>If k is a positive integer, which of the following must represent a even integer that is twice the value of an odd integer??</p>
<p>a 2k
b 2k+3
c 2k+4
d 4k+1
e 4k+2</p>
<p>Question 1 answer is 8.</p>
<p>Can anyone explain why??? and also for second question</p>
<p>Yea Question 1 is 8.</p>
<p>(x + 1)(x + h) = x^2 + 1x + hx + h
Since you know this is equivalent to x^2 + kx + 7, then h = 7, so
hx + x = k, or 7x + x = kx, so kx = 8x and k = 8. </p>
<p>Question 2, the answer is e, 4k + 2.
You can automatically eliminate answers a, b, and d.
Answer choice A is incorrect because the question asks for which of the following "must represent an even integer, twice the value of an odd integer". 2k / 2 could equal an odd or even integer.</p>
<p>Ex. if k = 5, then 2k = 10, and 10/2 = 5.</p>
<p>Answer choices b and d are incorrect because they represent odd integers. Any odd integer can be represented as 2x + y, where x is any integer, and y is an odd integer. (example: 9 can be represented as 2(4) + 1, or 2(3) + 3, etc.). Since the answer choices must be even integers, those two are incorrect.</p>
<p>This leaves you with answer choices c and e. Choice e is correct because when you divide it by 2, you are ALWAYS left with an odd integer.</p>
<p>(4k + 2) / 2 = 2k + 1.</p>
<p>As stated above, 2k + 1 would represent an odd integer.</p>
<p>Hope this helps.</p>
<p>what is exactly the difference between resentful and critical??</p>
<p>You can plug in 2 for k in question 2</p>
<p>I'm guessing you got a tone question on the CR wrong. There is a BIG difference when an author is being resentful in his writing and when he is being critical. </p>
<p>From the lovely creators of dictionary.com:</p>
<p>critical:
1. Inclined to judge severely and find fault.
2. Characterized by careful, exact evaluation and judgment
3. Of, relating to, or characteristic of critics or criticism</p>
<p>resentful:
1.Full of, characterized by, or inclined to feel indignant ill will.</p>
<p>Usually in more analytical writing, the author is critical. Resentful authors are easy to spot. They're the ones who are passionate about their loathing to the point of extremity.</p>
<p>I am lost on how h = 7. Can someone please describe this process further.</p>
<p>Thanks.</p>
<p>Well, this goes back to the distributive property. </p>
<p>If I have (x + 1)(x + 2), the product is: </p>
<p>(x * x) + (x * 2) + (1 * x) + (1 * 2)</p>
<p>So if x^2 + kx + 8 or something like that, the k must be the sum of the two numbers on the right of each thingy. </p>
<p>This is probably unclear, so I hope someone can explain it better, or better yet you can play around with the algebra and it will become clear. </p>
<p>The second question is really quite difficult to solve on theoretical grounds, but is pretty easy to brute force. Just imagine a bunch of values for k, and see if they meet the requirements. Most of those answer choices will get knocked off really quickly. </p>
<p>Theoretically, we can solve this problem by saying that the number must be even. </p>
<p>For the number to be even NO MATTER WHAT, it must have the number being multiplied by an even number, since any even * odd = even. So if any even number multiplied by k is odd, then an even number must be added to k, since adding an odd number would make the final answer odd. </p>
<p>So this brings us to: </p>
<p>even * k + even</p>
<p>Which brings us down to c and e. </p>
<p>The problem with c is that it involves adding 4 instead of adding 2. Every other even number divided by two is an odd number</p>
<p>2 yes
4 no
6 yes
8 no</p>
<p>So therefore the number produced by 2 or 4 times k will produce one of the "no" numbers, since 2 is in each of them. Adding 4 skips a step and takes you to another no, but adding 2 takes you to a yes number. </p>
<p>But that's all for kicks, actually thinking that out on the SAT will be a waste of time.</p>
<p>
[quote]
if k and h are constants and x^2+kx+7 is equivalent to (x+1)(x+h), what is the value of k??
[/quote]
Plug 0 for x:
0^2+k(0)+7 = (0+1)(0+h)
7 = h.
x^2+kx+7 = (x+1)(x+7).
That's right, k must be the sum of 1 and 7,
k = 1+7
k = 8.</p>
<p>If k is a positive integer, which of the following must represent a even integer that is twice the value of an odd integer??</p>
<p>a 2k
b 2k+3
c 2k+4
d 4k+1
e 4k+2</p>
<p>You might notice right away that
e 4k+2=2(2k+1), and 2k+1 is always an odd integer, so e is the answer.</p>
<p>You might also see that b and d are always odd. </p>
<h1>Even if you don't (that's kind of odd), it's not much longer.</h1>
<p>The brute force way:
Plug 1 for k (if we started with 2, it'd be faster).</p>
<p>a 2x1 = 2
b 2x1 + 3 = 5
c 2x1 + 4 = 6
d 4x1 + 1 = 5
e 4x1 + 2 = 6</p>
<p>b and d are out - they are not doubles of anything.
a, c and e are doubles of odd integers
2 = 2x1 (duh!)
6 = 2x3.</p>
<p>Plug 2 for k
a 2x2
c 2x2 + 4 = 8 = 2x4
e 4x2 + 2 = 10 = 2x5
That leaves e.</p>
<p>a theoretical approach for the second question:</p>
<p>no matter what k is, 2k + 1 must be odd (doubling k guarantees an even number, and adding one to an even number guarantees an odd number). doubling that odd number will result in an even number that is twice as much as an odd number, so: 2(2k + 1) = 4k + 2</p>
<p>^^
I thought I said the same thing:
[quote]
You might notice right away that
e 4k+2=2(2k+1), and 2k+1 is always an odd integer, so e is the answer.
[/quote]
</p>
<p>On the other hand, I did not explain why 2k+1 is odd.
Sorry, xitammarg: just feeling grouchy :eek: - it gets really crazy in the last few days leading to the test.</p>