<p>So, I found this question in a practice test (non CB).I know it can hardly appear on the real test, but still I will be very curious to know how to solve it .Here it is : </p>
<p>What is the value of the parameter m, for which the roots of the equation x^4-(3m+2)x^2 + m^2 = 0 are different and are part of an arithmetic sequence</p>
<p>Well… Ok, I’m gonna try doing this right now in this reply edit box. :D</p>
<p>Let p = x^2.</p>
<p>Then the equation is,</p>
<p>p^2 - (3m + 2)*p + m^2 = 0 ---- (1)</p>
<p>If the roots are a, b, then,</p>
<p>p^2 - (a + b)*p + ab = 0</p>
<p>p = a, b
x^2 = p</p>
<p>=> x = a, -a, b, -b</p>
<p>Let a < b</p>
<p>Then the sequence is,</p>
<p>-b, -a, a, b</p>
<p>If the common difference is d,</p>
<p>b - a = 2a = d
a = d/2</p>
<p>b - d/2 = d
b = 3d/2</p>
<p>Also, from eqn. 1</p>
<p>3m + 2 = a + b,
=> 3m = d/2 + 3d/2 - 2
=> m = (2d - 2)/3</p>
<p>Now we have m in terms of the common difference.</p>
<p>Hmm… Is that it?</p>
<p>x^4-(3m+2)x^2 + m^2 = 0</p>
<p>Two roots are:</p>
<p>a=((3m+2)+rt((3m+2)^2-4m^2))/2
b=((3m+2)-rt((3m+2)^2-4m^2))/2</p>
<p>So all roots are:
a, -a, b, -b</p>
<p>Assume a>b</p>
<p>Sequence is:
-a, -b, b, a</p>
<p>Common difference:
a-b = 2b</p>
<p>a=3b</p>
<p>((3m+2)+rt((3m+2)^2-4m^2))/2 = 3((3m+2)-rt((3m+2)^2-4m^2))/2</p>
<p>rt((3m+2)^2-4m^2) = 2(3m+2)-3rt((3m+2)^2-4m^2))</p>
<p>2(3m+2) = 4rt((3m+2)^2-4m^2))</p>
<p>4(3m+2)^2 = 16(3m+2)^2 - 64m^2</p>
<p>64m^2 - 12(3m+2)^2 = 0</p>
<p>64m^2 - 12(9m^2 + 12m +4) = 0</p>
<p>64m^2 - 108m^2 - 144m - 48 = 0</p>
<p>44m^2 + 144m + 48 = 0</p>
<p>11m^2 + 36m + 12 = 0</p>
<p>m=-0.377 or -2.90</p>
<p>What the… LOL! I have no idea if I’ve done this right. I did it all in the Quick Reply box so it could be completely wrong, plus it probably has a zillion silly mistakes.</p>
<p>One of two possible solutions:
m=6
x^4 - 20x^2 + 36 = 0
x={(-3)sqrt(2), -sqrt(2), sqrt(2), (3)sqrt(2)}</p>
<p>Here’s a better place for this type of questions:
<a href=“http://www.artofproblemsolving.com/Forum/index.php[/url]”>http://www.artofproblemsolving.com/Forum/index.php</a></p>
<p>Well it’s quite interesting, and I think we must think differently to solve this in about 1.5 minutes. Here is my solution:
Let p denote x^2, then: p^2 - (3m+2)p + m^2 = 0.
To get 4 different roots which form an arithmetic sequence, we must have 2 distinct roots of the above equation a and b, and they must satisfy the condition: a = b/9.
So when a=b/9, we have:
a+b = 10a = 3m+2 (1)
ab = 9a^2 = m^2
Hence: a = m/3 or a=-m/3.
Substitute a in (1):
<p>P.S: If this is a multiple choice question, the fastest way is to check each choice :D</p>
<p>x^4-(3m+2)x^2 + m^2 = 0 for reference.</p>
<p>So, we have the positive and negative of each root of x^2-(3m+2)x+m^2=0 to work with. Call the roots of the x^2 a and b, and say that b>a (without loss of generalization). Then b-a=2a (imagine that you have a number line</p>
<p>-b…-a…0…a…b</p>
<p>that would be an arithmetic sequence, right?)</p>
<p>Then, find a and b in terms of m (not sure how you would want to do that) and solve…I think it would work.</p>
<p>^Oh. I like your ps. :D</p>