Find the number of distinct real roots to the equation x 3 − 5 8 8 x = 5 5 0 0
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Very nice! @Matthew Christopher , I suggest, for your next problem, do the same thing for a quartic equation!
Real roots always exist as complements to pairs of complex numbers, right? So even though there can be 1, 2, 3 real roots, the remaining complex roots would always have to pair up. Hence 0 and 2 can never be the answer... Am I right, please correct if I am wrong.
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Yes. See fundamental theorem of algebra .
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@Pi Han Goh – @Pi Han Goh , I have heard that any cubic equation has at least one real root. So, doesn't it eliminate 0 also?
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@Vinayak Srivastava – @Vinayak Srivastava Please see my comment below. 0 is eliminated but not 2 as we are considering distinct roots.
True (we can only have 0 or 2 complex roots), so 0 can't be right - the blue line must always intersect the black one as the cubic has range y ∈ R . Looking at the graph, however, the blue line might be tangent to the black graph (at one of its stationary points), so we could have 2 distinct real roots, one of which is repeated. We could have 0 complex roots here. This means that 2 can be an answer.
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How can 2 be an answer? I read that complex roots always occur in conjugate pairs, so 2 can't be the number of real roots, only 1 or 3.
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@Vinayak Srivastava – Sure, they do. But if we have 2 distinct real roots then we have 0 complex roots. If you look at my sketch in the solution, clearly this is so because we can put the blue line tangent to the black one at a turning point. Really, there are 3 (not necessarily distinct) real solutions here, but that's not what my question asks.
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@Matthew Christopher – Oh, yes! Thanks for clearing my doubt! But I was just lucky, I didn't know anything what to do, just guessed by these outside things :)
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We could use the cubic discriminant , or instead, use the following trick by considering the intersection of the graphs of y = x 3 − 5 8 8 x and y = 5 5 0 0 .
Differentiating the left-hand side of the equation, we obtain 3 x 2 − 5 8 8 . If we set this to zero, the left-hand side cubic has stationary points where 3 x 2 − 5 8 8 ⟺ x 2 ⟺ x = 0 = 3 5 8 8 = 1 9 6 = ± 1 9 6 = ± 1 4
From the general shape of a cubic with a positive x 3 coefficient, the local maximum will be where x = − 1 4 (we could also consider the second derivative).
Here, x 3 − 5 8 8 x = ( − 1 4 ) 3 − 5 8 8 ⋅ ( − 1 4 ) = ( 1 4 ) 2 ( − 1 4 ) + 1 4 ( 1 4 ⋅ 4 2 ) = ( 1 4 ) 2 ( 4 2 − 1 4 ) = 1 9 6 ⋅ 2 8 = 5 4 8 8 .
Since 5 4 8 8 < 5 5 0 0 , we can sketch the situation like this:
So there is 1 distinct real solution (where the blue and grey lines intersect at x > 1 4 ).