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Let $a$ be largest real value of $x$ for which $x^3 - 8x^2 - 2x + 3 = 0$. Determine the integer closest to $a^2$.

How I tried to do this:

This is a third-degree polynomial, thus there are 3 positive/negative values of $x$. If I find the roots of this polynomial and take the largest one and square it, I should get $a^2$. I don't get the question's logic of "closest to $a^2$", why would I want to find that? How would I be able to find an integer "closest" to $a^2$? You either find it or you don't. I'd begin with testing small values, namely $0, -1, 1, 2, -2,$ etc. and then trying to find something that yields me an answer close to $0$. I'd then go on from there...

Thanks, would really appreciate help w/ this problem!

Felix Marin
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user164403
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  • Take its derivative and compute its two roots using the quadratic formula, to establish the intervals of monotony. Use the fact that $8^2=64<70<81=9^2$. – Lucian Sep 16 '14 at 23:18
  • I took it's derivative, $3x^2 - 16x - 2$, and then applied the quadratic formula, getting answers of 5.4, and -0.1222. I don't really understand what you mean by establishing the intervals of monotony. – user164403 Sep 17 '14 at 12:39
  • $a^2$is probably not an integer. The integer closest to $a^2$ is one of $\lfloor a^2\rfloor$, $\lceil a^2\rceil$. It is not unlikely that you could find this integer without computing $a$. –  Sep 18 '14 at 21:04

4 Answers4

3

Let $f(x)=x^3-8x^2-2x+3$. Note that $$f(-0.75)\lt 0\lt f(-0.6)$$ $$f(0.6)\lt 0\lt f(0.5)$$ $$f(8.16)\lt 0\lt f(8.21)$$ So, since we have $$8.16\lt a\lt 8.21\Rightarrow (66.5 \lt)\ 66.5856\lt a^2\lt 67.4041\ (\lt 67.5),$$ the integer closest to $a^2$ is $67$.

mathlove
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  • Quite a quick solution, though I do not fully understand your process. Could you provide an explanation? – user164403 Sep 16 '14 at 20:41
  • The three inequalities in my answer show that there are three real solutions $x_1,x_2,x_3$ such that $-0.75\lt x_1\lt -0.6,0.5\lt x_2\lt 0.6,8.16\lt x_3\lt 8.21$. Hence, we know $a=x_3$. Do you need more explanation? – mathlove Sep 16 '14 at 20:50
  • I kind of understand better now. Did you plug in values just to test, and then narrowed down your results? – user164403 Sep 16 '14 at 21:07
  • @user164403: well, I used wolfram alpha to know both three values and $a^2$. http://www.wolframalpha.com/input/?i=x%5E3-8x%5E2-2x%2B3%3D0 – mathlove Sep 16 '14 at 21:13
  • I'm doing contest questions and wolframalpha as well as calculators are not allowed. Is there any other way? – user164403 Sep 16 '14 at 21:28
  • Examine the graph of $y=f(x)$ by checking $f'(x)=0$. And plug in values to test, for example, $f(6),f(7),f(8),\cdots$ because the bigger root of $f'(x)=0$ is $x\approx 5.5$. – mathlove Sep 16 '14 at 21:35
1

If $a$ is a root of $x^3 -8x^2-2x+3$ then $a^2$ is a root of $x^3 - 68x^2+52x-9$. It's not hard to see that this is negative at 67 and positive at 68, so the root is somewhere in between those. Then testing 67.5 we can also see it's positive so it's closer to 67.

Nate
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We don't know that the polynomial has three roots. It could have one real and two complex roots. Without a calculator, I would note that the top two terms cancel at $x=8$, evaluate it there, and find $-13$. As the polynomial goes off to $+\infty$, there must be a root above $8$ As the first derivative is positive here, there is only one root greater than $8$. This will be $a$ and mathlove has shown that the closest integer to $a^2$ is $67$

Ross Millikan
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Looking at the plot of the polynomial, you see two roots in $[-1,1]$ and the largest one in $[8,9]$. So the requested integer is in range $[8^2,9^2]=[64,81]$.

Evaluating the polynomial at the square roots of these integers yields:

64  -13
65  -9.077761857
66  -5.061542103
67  -0.952069828
68  3.249942582
69  7.543798816
70  11.92881804
71  16.40433435
72  20.9696962
73  25.62426592
74  30.36741923
75  35.19854476
76  40.11704364
77  45.12232905
78  50.21382584
79  55.39097013
80  60.65320898
81  66

You can conclude.

For faster computation, you can use a dichotomic search, but that's another story.