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August 3rd, 2017, 03:46 PM   #1
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Equal roots of opposite direction

Hi
I have a question from a worksheet:

For what value(s) of k does the equation
3x^2 + (k - 2)x + k(k+1) = 0
have equal roots of opposite direction?

Many thanks
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August 3rd, 2017, 03:53 PM   #2
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If the roots are equal, they have the same "direction", whatever "direction" means.
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August 3rd, 2017, 04:04 PM   #3
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Don't quite understand as that is what appeared in the question...
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August 3rd, 2017, 04:26 PM   #4
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The question is misworded or mistyped. If k = 2, the roots are ±√2$\,i$.
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August 4th, 2017, 03:23 AM   #5
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My best guess (and is only a guess) is that the roots are intended to be "a" and "-a". Then the equation can be written as 3(x- a)(x+ a)= 3x^2- 3a^2= 0. Now compare that to the given form, 3x^2+ (k- 2)x+ k(k+1)= 0. We must have k- 2= 0 so k= 2.
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August 4th, 2017, 04:19 AM   #6
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Quote:
Originally Posted by Country Boy View Post
My best guess (and is only a guess) is that the roots are intended to be "a" and "-a". Then the equation can be written as 3(x- a)(x+ a)= 3x^2- 3a^2= 0. Now compare that to the given form, 3x^2+ (k- 2)x+ k(k+1)= 0. We must have k- 2= 0 so k= 2.
And if k = 2, k(k + 1) = 6 so

$6 = -\ 3a^2 \implies a^2 = -\ 2 \implies a = i\sqrt{2}$

as skipjack said. I guess that answer makes sense if the problem is talking about an Argand plane.
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