I can’t seem to get a handle on how to get started in this problem. Can someone give me a hint to get me started?
Find the value of c for which the roots of 2x^2 - 21x + c = 0 are in a ratio of 1:2.
General approach: Calculate the roots abstractly (in terms of c). Figure out their ratio (again, you’ll get something in terms of c). Then “invert” this to get c in terms of the ratio. Then substitute in the particular ratio of 1:2, and you’ll have your answer for what c should be.
If you need more help, there’s a formula you’ve probably been told to commit to rote memory for solving equations like ax^2 + bx + c = 0…
Take a step back, how would you find the roots if c=12?
Just to be clear, the problem is:
Find the roots (x,c) of 2x[sup]2[/sup] - 21x + c = 0 where 2x = c
Well, when you see c, replace with 2x. They’re equal, after all.
Thanks Bryan. When I replace c with 2x I solve it with 2 = 9.5, c =19. I never would have gotten those numbers without your hint.
I don’t think that’s what the problem is, Bryan Ekers.
I think the problem is:
Find a c such that there exists an x such that 2x^2 - 21x + c = 0 and 2(2x)^2 - 21(2x) + c = 0.
What leads you to say the ratio 1:2 is between one of the roots and c, rather than between the two roots?
The problem says the roots are in a ratio of 1:2, not that x and c are in a ratio of 1:2.
I’m reading the problem the same as Santo Rugger & Indistinguishable, and solving that gives a nice solution without irrational numbers as well.
ETA: Actually, there are 2 solutions for c, but one of them is degenerate.
That’s definitely not the right answer to the original problem. Your quadratic can be factored as (x - r[sub]1[/sub])(x - r[sub]2[/sub]), where r[sub]1[/sub] and r[sub]2[/sub] depend on the value of c. You’re being asked to find the value of c such that r[sub]1[/sub]/r[sub]2[/sub] = 2.
Unless there’s a convention that I’m not familiar with, there’s a slight ambiguity in the problem. If r[sub]1[/sub] < r[sub]2[/sub], solving r[sub]1[/sub]/r[sub]2[/sub] = 2 and r[sub]1[/sub]/r[sub]2[/sub] = 1/2 for c will give you different answers, but both equations specify that the roots are in a 1:2 ratio.
[spoiler]
2x^2 - 21x + c = 0
roots = (-b +/- (b[sup]2[/sup] -4ac)[sup]1/2[/sup]) / 2a
= -21 +/- (441 - 8c)[sup]1/2[/sup]) / 4
Therefore,
-21 + (441 - 8c)[sup]1/2[/sup]) / 4 = 2 * (-21 - (441 - 8c)[sup]1/2[/sup]) / 4)
which can be rewritten as
-21 + (441 - 8c)[sup]1/2[/sup]) = 2 * (-21 - (441 - 8c)[sup]1/2[/sup]))
OR
-21 - (441 - 8c)[sup]1/2[/sup]) / 4 = 2 * (-21 + (441 - 8c)[sup]1/2[/sup]) / 4)
which can be rewritten as
-21 - (441 - 8c)[sup]1/2[/sup]) = 2 * (-21 + (441 - 8c)[sup]1/2[/sup]))[/spoiler]
Solve for c.
I’m getting 441/8 and 1323/32.
You’ve misnegated your b. There is a very clean answer for c.
Santo you made an error:
In the first step b = -21, so -b = 21.
Without redoing all your math, I got c = 49, and c = 0. Both of which work. If c = 49, then x = 7, 3.5. If c = 0, then x = 0, 0.
I had previously written “And I’m not sure what ultrafilter means about ambiguity. As I see it, there is a very clean unique value of c with the specified property; the ambiguity only manifests as the usual business about having both positive and negative square roots in the discriminant, but this distinction is erased when recovering c.”, but sheepishly erased it on seeing muldoonthief’s post. But now I realize what has happened. You highlight an interesting point, but I wouldn’t actually technically count the second solution you gave, on the grounds that the root you’re using on both sides of the ratio is not actually a double root (there’s another additional root in that case, so you are exhibiting 1:2 not as a ratio between the two roots, but just between one root and itself).
Hence my “just to be clear” bit, which I guess should have been formed more as a question. By “roots” I thought the OP meant the variables in the equation, x and c. If instead he means:
“Find a value for the constant c, such that it leads to two roots, x[sub]1[/sub] and x[sub]2[/sub], where x[sub]1[/sub] = 2x[sub]2[/sub]”
…then that’s a whole other problem, but not difficult.
On recomputation (and without reading the spoiler boxes of others), I figure:
[spoiler]c = 49
2x[sup]2[/sup] - 21x + 49 = 0
x[sub]1[/sub] = 7
x[sub]2[/sub] = 3.5[/spoiler]
Indistinguishable, I agree, the second answer isn’t an interesting one (you can get it from inspection), but I’d solve it both ways and present both in my work, just for completeness sake/I was a smartass in school.
It’s not that the second answer isn’t interesting; I think the second answer is interesting. But I don’t think it’s a correct answer to the problem. What’s interesting about it is the subtle distinction it elucidates between the problem actually posed and a nearly, but not quite, equivalent formulation (such as the one I erroneously gave in post #6). Specifically,
When c = 0, the two roots are 0 and 21/2. These are not in the ratio 1:2, so this is not a correct solution. It is true that 0 and itself are in the ratio 1:2 (and every other ratio as well), but that’s not really here or there…
Here’s a start:
Assume that one of the roots is w. Then the other root is 2w. So the equation can be written like this (x-w) (x-2w) = 0.
That’s a good suggestion, leading to a straightforward solution
2*(x-w)*(x-2w)=2x^2-21x+c
2x^2-6wx+4w^2=2x^2-21x+c
6w=21 which means w=7/2 (and 2w=7)
and c=4w^2 which means c=49.
Nice