couple act problems

Test Preparation
1

<p>2x^2 + 6x = 36, what are the possible values of x?</p>

<p>f. -12 and 3
g. -6 and 3
h. -3 and 6
j. -3 and 12
k. 12 and 15.</p>

<p>i tried substituting those numbers for x and none of them come out 36. how??</p>

<li>i^2 = -1, (4 + i)^2 = ?</li>
</ol>

<p>f. 15
g. 17
h. 15 + 4i
j. 15 + 8i
k. 16 + 4i</p>

<p>can you show me step by step on how you wold arrive to that answer? if i were guessing i would eliminate f, g, & k and guess H…now looking at the answer sheet the answer is J. please explain</p>

<li>what is the value of log(subscript 2)8?
how would you punch this in the calculator - i havent taken trig yet.</li>
</ol>

2

<ol>
<li>See</a> here</li>
<li>See</a> here</li>
<li>See</a> here</li>
</ol>

<p>Easiest solution: Get a TI-83 Plus (I'd recommend a TI-89 Titanium but ACT doesn't allow it) and learn how to use it.</p>

3

<p>You can solve that in like 5 seconds if you have QuadForm on your TI-83. Just a tip.</p>

4

<p>g and j are the answers. There's an i-button on your 83 and you can graph the first problem and find the zeros. Still, sometimes it's just better to solve it algebraically.</p>

5

<p>it's sort of inefficient to use the quadratic for such a simple question. divide the entire equation by 2, which leaves you with x^2 + 3x - 18 = 0. factor: (x + 6)(x - 3) = 0 ; x = -6, x = 3</p>

6

<p>
[quote]

it's sort of inefficient to use the quadratic for such a simple question. divide the entire equation by 2, which leaves you with x^2 + 3x - 18 = 0. factor: (x + 6)(x - 3) = 0 ; x = -6, x = 3

[/quote]

I guess you're right, but the quadratic formula is a simple, reliable method to find the zeros of most second degree polynomial equations.</p>