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If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$

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    $\begingroup$ hi, have you tried anything already? If so it would be useful to share what you have done so far. $\endgroup$
    – Joe Tait
    Jun 28 '13 at 13:14
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Because $x^2+3x+1=0$, we have $x^2=-3x-1$ and also $x^2+2x+1=-x$, for $x=a,b$. Hence $$\left(\frac{a}{b+1}\right)^2=\frac{a^2}{(b+1)^2}=\frac{-3a-1}{-b}=\frac{3a+1}{b}$$ By symmetry, the desired expression is $$\frac{3a+1}{b}+\frac{3b+1}{a}=\frac{3a^2+a}{ab}+\frac{3b^2+b}{ab}=\frac{3a^2+a+3b^2+b}{ab}=\frac{3(-3a-1)+a+3(-3b-1)+b}{ab}=\frac{-8(a+b)-6}{ab}$$ Lastly, because $a,b$ are roots of $x^2+3x+1$, we know that $ab=1$ and $a+b=-3$. Plugging this into our final expression gives $$\frac{-8(-3)-6}{1}=18$$

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  • $\begingroup$ At least assuming that $a \not= b$. When this is not the case, the expression is equal to $3 + \sqrt{5}$. $\endgroup$
    – fuglede
    Jun 28 '13 at 13:21
  • $\begingroup$ @vadim123: OP only assumed that $a$ and $b$ were roots; not that they were distinct. $\endgroup$
    – fuglede
    Jun 28 '13 at 13:24
  • $\begingroup$ Add.: The comment my last comment replied to is deleted so it doesn't really make sense any more. $\endgroup$
    – fuglede
    Jun 28 '13 at 13:36
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Hint: $(a+1)^2 + a = 0$ and so $\frac{1}{(a+1)^2} = -\frac{1}{a}$

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We note that the expression given is a symmetric rational function of the roots of the original polynomial, so can be expressed in terms of the coefficients of the original polynomial. We then use the basic facts we know about the sum and product of the roots, and the fact that both $a$ and $b$ satisfy the polynomial to achieve successive simplification of otherwise unwieldy terms. Here are some hints for a way through.

Note that $a+b=-3$ and $ab=1$ so that $(a+1)(b+1)=ab+(a+b)+1=-3+2=-1$

Use this to put the whole thing over a common denominator and simplify.

Note also that $a^2(a+1)^2=a^2(a^2+2a+1)=a^2(a^2+3a+1-a)=-a^3$

You might also need to use $a^2+b^2=(a+b)^2-2ab$


Since the full answer closest in spirit to this has been deleted, note first $(a^2+b^2) = (-3)^2-2=7$. Then put the desired expression over common denominator $(a+1)^2(b+1)^2=1$ to obtain:$$a^2(a+1)^2+b^2(b+1)^2=-a^3-b^3=3(a^2+b^2)+(a+b)=21-3=18$$ [see comment for middle step]

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  • $\begingroup$ Use also $-a^3=3a^2+a$ obtained from $a(a^2+3a+1)=0$ $\endgroup$ Jun 28 '13 at 13:25
  • $\begingroup$ @MarkBennet:why is $ab=1$? $\endgroup$
    – Alex
    Jun 28 '13 at 21:24
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    $\begingroup$ @Alex $(x-a)(x-b)=x^2+3x+1$ - expand and equate coefficients. This works because for $x=a$ or $x=b$ we have $(x-a)(x-b)=0$ - it is east to show that there is a unique monic quadratic equation which has roots $a$ and $b$. $\endgroup$ Jun 29 '13 at 3:07
  • $\begingroup$ @MarkBennet: thanks, that's great! $\endgroup$
    – Alex
    Jun 29 '13 at 3:36
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$$a+b=-3,ab=1$$ $$(a+b)^2=a^2+b^2+2ab\implies a^2+b^2=7$$ $$(a+b)^3=a^3+b^3+3ab(a+b)\implies a^3+b^3=-18$$ $$(a^2+b^2)^2=a^4+b^4+2(ab)^2\implies a^4+b^4=47$$

$${\left(\dfrac {a}{b+1}\right)}^2+{\left(\dfrac {b}{a+1}\right)}^2$$ $$\dfrac {a^2(a+1)^2+b^2(b+1)^2}{((b+1)(a+1))^2}$$ $$\dfrac {a^4+2a^3+a^2+b^4+2b^3+b^2}{(ab+a+b+1)^2}$$ $$\dfrac {a^4+b^4+2(a^3+b^3)+a^2+b^2}{(1-3+1)^2}$$ $${47+2(-18)+7}$$ $$18$$

There is one alternate as in comment @mark suggest: $$\dfrac {a^4+2a^3+a^2+b^4+2b^3+b^2}{(ab+a+b+1)^2}$$ $${a^4+3a^3+a^2-a^3+b^4+3b^3+b^2-b^3}$$ $${a^2(a^2+3a+1)-a^3+b^2(b^2+3b+1)-b^3}$$ since $a^2+3a+1=b^2+3b+10=0$ $$-a^3-b^3$$ $$18$$

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  • $\begingroup$ @vadim123 I've edited my answer. I think now I'm clear to you. $\endgroup$ Jun 28 '13 at 13:53
  • $\begingroup$ Note that you can reduce the numerator to $-a^3-b^3$ quite easily using the original equation. $\endgroup$ Jun 28 '13 at 14:55
  • $\begingroup$ @MarkBennet in which line? $\endgroup$ Jun 28 '13 at 16:23
  • $\begingroup$ $a^4+2a^3+a^2=a^4+3a^3+a^2-a^3=a^2(a^2+3a+1)-a^3=a^3$ and the same for $b$ - 7th line I think. But it is also neat to know that you can calculate the higher powers if need be. $\endgroup$ Jun 28 '13 at 16:29
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HINT:

As $ab=1, \frac a{b+1}=\frac{a}{\frac1a+1}=\frac{a^2}{a+1}=y$(say)

So, $a^2=y(a+1)$

and again $a^2+3a+1=0\implies a^2=-3a-1$

So, $ay+y=-3a-1\implies a=-\frac{y+1}{y+3} $

As $a$ is root of the given eqaution $$\left(-\frac{y+1}{y+3} \right)^2+3\left(-\frac{y+1}{y+3} \right)+1=0$$

Simply to get $y^2+4y-1=0$

Using Vieta's Formula, the required sum will be $(-4)^2-2(-1)=18$

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From the content of this thread follows,

$x^2 + 3x +1 = 0 \Leftrightarrow x^2 +2x+1=-x \Leftrightarrow \boxed{(x+1)^2 =-x} (*)$

$\frac{a^2}{(b+1)^2}+\frac{b^2}{(a+1)^2} \Leftrightarrow \left(\frac{b^2}{b^2}\right)\frac{a^2}{(b+1)^2}+\left(\frac{a^2}{a^2}\right)\frac{b^2}{(a+1)^2} \\ \hspace{3.05cm}\Leftrightarrow \underset{\scriptsize -1 \, by \, (*)}{\boxed{\frac{b}{(b+1)^2}}}\frac{a^2 b}{b^2} + \underset{\scriptsize -1}{\boxed{\frac{a}{(a+1)^2}}} \frac{b^2 a}{a^2} \\ \hspace{3.05cm}\Leftrightarrow -\frac{a^2}{b}-\frac{b^2}{a}\\ \hspace{3.05cm}\Leftrightarrow \frac{- a^3 - b^3}{ab} \\ \hspace{3.05cm}\Leftrightarrow \boxed{- a^3 - b^3}$

$-a^3-b^3 = -(a+b)(a^2+b^2-ab+2ab-2ab) \\ \hspace{1.85cm} = 3((a+b)^2-3ab)\\ \hspace{1.85cm} = 3(9-3) \\ \hspace{1.85cm} = \boxed{18} $

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