1. ## SAT Question

1. If k and h are constants and x^2 + kx + 7 is equivalent to (x+1)(x+h), what is the value of k?

(a) 0
(b) 1
(c) 7
(d) 8

2. If k is a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?
(a) 2k
(b) 2k+3
(c) 2k+4
(d) 4k+1
(e) 4k+2

i thought that the answer was (a) but the actual answer is (e) and i don't know why

3. If x and y are positive integers and 4(2^x) = 2^y, what is x in terms of y
(a) y-2
(b) y-1
(c) y
(d) y+1
(e) y+2

i didn't know how to solve this because what i did was:
8^x = 2^ y
2^3x=2^y
3x=y
x=y/3

but there was no option for that..

4. If n/(n-1) x 1/n x n/(n+1) = 5/k for positive integers n and k, what is the value of k?
(a) 1
(b) 5
(c) 24
(d) 25
(e) 26

i simplified the whole thing down to:
k = 5n + 5/n
and i do not know what to do next
thank you so much!

2. Originally Posted by juliak
1. If k and h are constants and x^2 + kx + 7 is equivalent to (x+1)(x+h), what is the value of k?

(a) 0
(b) 1
(c) 7
(d) 8
Since this is SAT, substitute all the options and get the answer. -1 must be a root of x^2 + kx + 7.
(a) x^2 + 7, -1 is not a root
(b) x^2 + x + 7, -1 is not a root
(c) x^2 + 7x + 7, -1 is not a root
(d) x^2 + 8x + 7, -1 is a root

Direct method is to compare (x+1)(x+h) with x^2 + kx+7.
x^2 + (h+1)x + h = x^2 + kx + 7.
So h+1 = k and h = 7;

Thus k = 7+1 = 8;

Originally Posted by juliak
2. If k is a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?
(a) 2k
(b) 2k+3
(c) 2k+4
(d) 4k+1
(e) 4k+2
i thought that the answer was (a) but the actual answer is (e) and i don't know why
thank you so much!
(a) is not the answer. if k is 2, then 2k is not "twice the value of an odd integer"

However 2k+1 is always odd and therefore 2(2k+1) = 4k+2 is "twice the value of an odd integer". So (e) is correct

3. Thank you very much but I am a bit confused about the direct method - how did you know that h was 7?

4. Originally Posted by juliak

3. If x and y are positive integers and 4(2^x) = 2^y, what is x in terms of y
(a) y-2
(b) y-1
(c) y
(d) y+1
(e) y+2

i didn't know how to solve this because what i did was:
8^x = 2^ y
2^3x=2^y
3x=y
x=y/3

but there was no option for that..
Note that $4(2^x) \ne 8^x$

$4(2^x) = 2^2 \cdot 2^x = 2^{x+2}$

$2^{x+2} = 2^y$

$x + 2 = y$

$x = y - 2$ Answer A

4. If n/(n-1) x 1/n x n/(n+1) = 5/k for positive integers n and k, what is the value of k?
(a) 1
(b) 5
(c) 24
(d) 25
(e) 26

i simplified the whole thing down to:
k = 5n + 5/n
and i do not know what to do next
$
\frac{\not{n}}{n-1} \cdot \frac{1}{\not{n}} \cdot \frac{n}{n+1} = \frac{5}{k}
$

$
\frac{n}{(n-1)(n+1)} = \frac{5}{k}$

$kn = 5(n^2 - 1)$

$5n^2 - kn - 5 = 0$

Your hint here is that k and n are integers. This means the discriminant of this quadratic equation $\Delta = b^2 - 4(a)(c)$ must be a perfect square. Otherwise n will not be an integer.

For this equation,
$\Delta = (-k)^2 - 4(5)(-5) \Rightarrow k^2 + 100$

So $k^2 + 100$ is a perfect square. Substitute the choices to see which value of k works.

$a) 1^2 + 100 = 101$ not a perfect square.
$b) 5^2 + 100 = 125$ not a perfect square.
$c) 24^2 + 100 = 676 = 26^2$ This is a perfect square. Answer C
$d) 25^2 + 100 = 725$ not a perfect square.
$e) 26^2 + 100 = 776$ not a perfect square.

Thank you very much but I am a bit confused about the direct method - how did you know that h was 7?
Basically you compare the left hand side of the equation with the right hand side.

$x^2 + (h+1)x + h = x^2 + kx + 7$
Comparing $x^2 \Rightarrow x^2 = x^2$
Comparing $x \Rightarrow (h+1)x = (k)x \Rightarrow h+1 = k$
Comparing constants $\Rightarrow h = 7$