Math question help (again)?

Test Preparation SAT Preparation
1

<p>Point Q is the vertex of an 8-sided polygon with 8 sides of equal length and 8 equal angles. When all possible diagonals are drawn from point Q in the polygon, how many triangles are formed?
A. 4
B. 5
C. 6
D. 7
E. 8</p>

<p>I draw an octogon and it seemed overly straightforward: E. 8 triangles. But the number of triangles is C. 6. Any ideas? I'm either just geometrically impaired or diagonal doesn't mean a line from the vertex to a point between two edges...</p>

<hr>

<p>In the equation below, k and m are constants. If the equation is true for all values of x, what is the value of m?
(x-8)(x-k) = x^2 - 5kx + m
A. 8
B. 16
C. 24
D. 32
E. 40
Answer: B. 16. NO idea WHATSOEVER as to how to solve this. I honestly just simplified the factored out thing but that all that allowed me to do was cancel out the x squares. Any ideas?</p>

<p>All help is well appreciated!</p>

2

<p>For the first question, if you draw out an octagon and pick one vertex and draw lines from that vertex to each of the other ones, except for the two adjacent to it because those would not form triangles, there should be 5 different lines drawn, forming 6 different triangles</p>

3

<p>For the first one, when you drew your triangles, did you use only the vertices of the given ocatgon? That DOES give 6 triangles (as in the formula (n-2)x180 for the total of the angle measures). Or did you introduce a new vertex at the center? That would give you 8 triangles, but they didn’t mean for you to stick a new vertex in there…</p>

<p>As for the second one, it has been discussed here many times, including here:</p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/759503-2-math-problems-stumped-usual-800er.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/759503-2-math-problems-stumped-usual-800er.html&lt;/a&gt;&lt;/p&gt;

<p>But if you don’t like the solutions in that thread, google it – there are at least 6 other threads to choose from :)</p>

4

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<p>This may make me sound really, really idiotic, but I mistook ‘vertex’ as the point in the center of the polygon, not vertex as in a point in between two edges. </p>

<p>I am baffled at my stupidity…</p>

5

<p>For the second one, start by multiplying out the quantities. So x^2-kx-8x+8k=x^2-5kx+m. Noticing that the left side has a common x to two terms, -kx and -8x, we can simplify that (and cancelling the x^2 terms) to (-k-8)x+8k = -5kx + m. Since x is a variable, -k-8 = -5k -> -8=-4k -> k = 2. Then for the constant terms, 8k and m, they are equal so m = 8k = 8(2) = 16 or B Q.E.D</p>

<p>If you’re confused why -k-8 = -5k and why m=8k, think of the the two equations as being equal but simply in a different form. x is a variable and can be any number, so in order for the two equations to be equal, the coefficient of x and the constant term (when x=0, 8k and m) must be equal for the two equations.</p>

6

<p>Here’s another solution to the second one that is quicker (thus good for the SAT), but requires a little more technical knowledge.</p>

<p>Recall: In the quadratic equation x^2+bx+c=0, c is the product of the roots, and -b is the sum of the roots.</p>

<p>Now, the roots of (x-8)(x-k) are 8 and k. </p>

<p>So 8+k=5k. Thus 8=-4k and k=2. </p>

<p>So m=8k=8*2=16, choice (B).</p>

7

<p>Dr. Steve, how would you work this one then?</p>

<p>(6x+y)(x-h) = 6x^2 +5x-6. What is the value of y?</p>

8

<p>This is a hard question. I’m not sure I would use the method I described above for this, but if you want to you need the more general result:</p>

<p>In the quadratic equation ax^2+bx+c=0, the product of the roots is c/a, and the sum of the roots is -b/a.</p>

<p>The roots of (6x+y)(x-h) are -y/6 and h.</p>

<p>So h-y/6=-5/6, and -hy/6=-6/6. Equivalently, y-6h=5 and hy=6.</p>

<p>So we get a pretty difficult system of equations (this is the same system you would get if you equated coefficients).</p>

<p>I would solve this by first multiplying the second equation by 6, and then substituting 6h for y-5. From the first equation:</p>

<p>6hy=36, so (y-5)y=36. Thus y^2-5y-36=0. The left factors: (y-9)(y+4)=0. So y=9 or y=-4. We get two solutions.</p>

<p>This couldn’t possibly be an SAT problem.</p>

9

<p>thank you both!</p>

10

<p>As DrSteve stated, a general method may not be the best way to solve special kinds of equations.
For the quadratic equation ax^2+bx+c=0 with the rooits x1 and x2
ax^2+bx+c=a(x-x1)(x-x2)
For 6x^2+5x-6=0 x1=-3/2 and x2=2/3
6x^2+5x-6=6(x+3/2)(x-2/3)</p>

<p>Distributing 6 in two different ways
6x^2+5x-6=(6x+9)(x-2/3) or
6x^2+5x-6=(6x-4)(x+3/2).</p>

<p>(6x+y)(x-h)=(6x+9)(x-2/3), y=9
(6x+y)(x-h)=(6x-4)(x+3/2), y=-4</p>

<p>This question definitely belongs to the Subject Math Level 2.</p>

11

<p>@marsha.ss - you are quite welcome!
Sorry I corrupted the order of posts by deleting and then reposting my solution.</p>