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 How many roots? (Posted on 2011-07-31)
Prove that the equation
x^4-4*x-1=0
has at most 2 real number solutions.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 re: Solution (spoiler) - I think | Comment 2 of 6 |
(In reply to Solution (spoiler) - I think by Kenny M)

Good thinking and a nice approach, Kenny.

But there is a conceptual error. By definition of a+-bi and c+-di, the coefficients a, b, c, d are all real numbers. Therefore, the conclusion that 2 of them must be imaginary is, ipso facto, incorrect.

Here's a slightly different take on the problem:

if f(x) = x^4 - 4x -1, then f'(x) = 4x^3 - 4

So, f'(x) = 0 iff x = +1  => the curve representing the polynomial has only 1 turning point at x = +1.

Now f(-1) = +4  and  f(1) = -4 and f(2) = +7

Therefore, the curve is decreasing to the left of x = +1 and increasdng to the right of x = +1.

So, it is concave upwards, and must cross the x-axis at 2 points (i.e.., real numbers) between x = -1 and x = +2.

So, there are precisely 2 real roots.

Edited on August 4, 2011, 8:05 pm
 Posted by JayDeeKay on 2011-08-04 20:04:19

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