Roots of a cubic equation

Calculus Level 2

Find the number of distinct real roots to the equation x 3 588 x = 5500 x^3-588x=5500

2 1 3 0

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1 solution

We could use the cubic discriminant , or instead, use the following trick by considering the intersection of the graphs of y = x 3 588 x y=x^3-588x and y = 5500 \color{#4200ff}{y=5500} .

Differentiating the left-hand side of the equation, we obtain 3 x 2 588 3x^2-588 . If we set this to zero, the left-hand side cubic has stationary points where 3 x 2 588 = 0 x 2 = 588 3 = 196 x = ± 196 = ± 14 \begin{aligned}3x^2-588&=0\\\iff x^2&=\frac{588}{3}=196\\\iff x&=\pm\sqrt{196}\\&=\pm 14\end{aligned}

From the general shape of a cubic with a positive x 3 x^3 coefficient, the local maximum will be where x = 14 x=-14 (we could also consider the second derivative).

Here, x 3 588 x = ( 14 ) 3 588 ( 14 ) = ( 14 ) 2 ( 14 ) + 14 ( 14 42 ) = ( 14 ) 2 ( 42 14 ) = 196 28 = 5488 x^3-588x=(-14)^3-588\cdot (-14)=(14)^2(-14)+14(14\cdot 42)=(14)^2(42-14)=196\cdot 28=\color{#EC7300}{\boxed{5488}} .

Since 5488 < 5500 5488<5500 , we can sketch the situation like this:

So there is 1 \color{#20A900}{\boxed{1}} distinct real solution (where the blue and grey lines intersect at x > 14 x>14 ).

Very nice! @Matthew Christopher , I suggest, for your next problem, do the same thing for a quartic equation!

Yajat Shamji - 10 months, 2 weeks ago

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Good idea!

Matthew Christopher - 10 months, 2 weeks ago

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.

Mahdi Raza - 10 months, 1 week ago

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Yes. See fundamental theorem of algebra .

Pi Han Goh - 10 months, 1 week ago

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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?

Vinayak Srivastava - 9 months, 4 weeks ago

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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.

Matthew Christopher - 9 months, 4 weeks ago

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 y\in\mathbb{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.

Matthew Christopher - 10 months, 1 week ago

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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.

Vinayak Srivastava - 9 months, 4 weeks ago

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@Vinayak Srivastava Sure, they do. But if we have 2 2 distinct real roots then we have 0 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 3 (not necessarily distinct) real solutions here, but that's not what my question asks.

Matthew Christopher - 9 months, 4 weeks ago

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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 :)

Vinayak Srivastava - 9 months, 4 weeks ago

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