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Let a and b be the roots of this equation:

x^2 - x - 5 = 0

Find the value of

(a^2 + 4b - 1)(b^2 + 4a - 1)

Without calculating the values of a and b.

I saw this on a problems site and tried it but I got 100 and I don't think that's correct.

3 Answers 3

5

Hint:

$$ a^2 + 4b - 1 = a^2 - a - 5 + 4b + a + 4 = 4b + a + 4 $$

11

Make repeated use of $a^2=a+5$, $b^2=b+5$, $a+b=1$, $ab=-5$.

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    How do we know these? Could you explain the first two?2011-06-06
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    @Zinger, what does it mean for $a$ to be a root of $x^2-x-5=0$?2011-06-06
11

This is an exercise in elementary symmetric polynomials. Write $s_1=a+b$ and $s_2=ab$. From the equation $$ x^2-x-5=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-s_1x+s_2 $$ we can read that $s_1=1$ and $s_2=-5$.

The number that you wanted to know is $$ (a^2+4b-1)(b^2+4a-1)=a^2b^2+4(a^3+b^3)+16ab-4(a+b)-(a^2+b^2)+1. $$ We need to express the quantities in parenthesis in terms of $s_1$ and $s_2$. This is not too difficult, because from $$ s_1^2=a^2+2ab+b^2=(a^2+b^2)+2s_2 $$ we get $a^2+b^2=s_1^2-2s_2=11$. Similarly from $$ s_1^3=a^3+3a^2b+3ab^2+b^3=(a^3+b^3)+3ab(a+b) $$ we get that $a^3+b^3=s_1^3-3s_1s_2=16.$

Putting all this together gives $$ (a^2+4b-1)(b^2+4a-1)=25+64-80-4-11+1=-5. $$

What makes this tick is that $(a^2+4b-1)(b^2+4a-1)$ is symmetric in the unknowns $a$ and $b$. IOW if you swap the values of $a$ and $b$ nothing will change. Such polynomial functions can always be written in terms of the elementary symmetric polynomials $s_1$ and $s_2$. This result can be generalized to several unknowns. The buzzword "elementary/basic symmetric polynomials" should give you enough material. The power sums such as $a^3+b^3$ are a well-known special case. The buzzword "Newton's identities" helps you there.

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    Given that I neglected to take advantage of the starting equation to reduce the degrees of the polynomials, an upvote feels a bit generous. Thanks, anyway :-)2011-06-03
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    Your answer not only resolved the question at hand, but provided further background for the OP. +1 for that alone!2011-06-03
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    Answer on a PSQ.2018-12-08
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    @close-deleteautomaton Indeed. Back in 2011 the standards were different. IIRC this was my first or second day on MSE :-)2018-12-09
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    @close-deleteautomaton IIRC this was my first post on MSE. It received a downvote (retracted years later). That downvote played a role in cementing my anti-PSQ stance. Thanks for reminding me of why it is important not to do askers' homework for them :-) Also observe that even back in the day I also try to "teach a man to fish" as opposed to just "give them a fish".2018-12-09
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    @JyrkiLahtonen hopefully it added a little perspective to look back.2018-12-09
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    Incidentally, I don't think the question is that harmful.2018-12-09
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    @close-deleteautomaton Back in 2011 pretty much every post had the benefit of not excessively duplicating earlier material. That. Has changed since. Of course, "new content" should not be the only criterion in deciding acceptability.2018-12-09
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    @JyrkiLahtonen I favour use of the duplicate content functionality. It works well. I imagine it is difficult not to be jaded by seeing same questions repeated year after year but to others these are new questions.2018-12-09