# Solve $5a^2 - 4ab - b^2 + 9 = 0$, $- 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$

Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$

I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.

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 homework should not be used as a standalone tag; see tag-wiki and meta. – Martin Sleziak Oct 10 '12 at 15:49

You can notice that many terms of $$(b+5a-4)^2=b^2+10ab-8b-40a+25a^2+16$$ appear in the first equation. Similarly, in the first one, you can notice $(b+2a)^2$.

By algebraic manipulation you get that the original equations are equivalent to \begin{align} (b+5a-4)^2&=4(a^2+1)\\ (b+2a)^2&=9(a^2+1) \end{align} which implies $4(b+2a)^2=9(b+5a-4)^2$ and $2(b+2a)=\pm 3(b+5a-4)$. This should simplify things a little. (In each of the two possibilities you can express $b$ using $a$ as a linear expression. Then you will get a quadratic equation in $a$. Or you can start by eliminating $a$.)

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 I've corrected $(b+2)^a$. Are there other typos? – Martin Sleziak Oct 10 '12 at 15:47

Subtract the two equation (thereby eliminating the $b^2$ term) and solve for $b$, which you can plug into either equation and get an equation in $a$; so $a$ is a solution to $(8a^2-24a+15)(3+4a)^2$. The roots are $a=-3/4, 3/2\pm \sqrt{6}/4$.

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By arrange the terms, you can note that

$$\begin{cases} 5a^{2}-4ab-b^2+9=0\\ -21a^{2}-10ab+40a-b^{2}+8b-12=0 \end{cases} \Leftrightarrow \begin{cases} 9(a^{2}+1)=(2a+b)^{2}\\ 4(a^{2}+1)=(5a+b-4)^{2} \end{cases}$$

So we get

$$\begin{cases} 9(a^{2}+1)=(2a+b)^{2}\\ 4(2a+b)^{2}=9(5a+b-4)^{2} \end{cases}$$

According to the second equation, we can get two cases of first relation between $a$ and $b$, then substitute them into the first equation respectively, we can get all the cases of values of pairs $(a,b)$.

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Note that \begin{equation*} 4(5a^2 - 4ab - b^2 + 9) - 9(-21a^2-10ab+40a-b^2+8b-12) = (19a+5b-12)(11a+b-12). \end{equation*}

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 How did you note that ? – Belgi Oct 11 '12 at 6:11

$21a^2+10ab+b^2-10a-8b+12=0$

$(7a+b)(3a+b)-4(7a+b+3a+b)+12=0$

$(7a+b-4)(3a+b-4)=4--->(1)$

$5a^2-4ab-b^2=(5a+b)(a-b)=-9--->(2)$

If $7a+b-4=\frac 2 k, 3a+b-4=2k$,

Express $a,b$ in terms of $k,$ replace their values in (2).

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According to maple, $a=-3/4, b=21/4$ is one solution, and if $r$ is a solution of $2x^2-12x+15=0$ then $a=r/2, b=12-(11/2)r$ is another solution. As the quadratic here has two real zeros, there are three pairs $(a,b)$ of real numbers for your system.

Of course this is not an elegant solution, and even worse it relies on maple. But it seemed odd to me since when the equations are subtracted and the result solved for $b$, we get $b=(26a^2-40a+21)/(8-6a)$, which may then be put into the first equation, so I would have expected a fourth degree equation for a.

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I admit the above is not elegant. But I thought it interesting that, if you subtract the two equations and solve for b, and then plug that into the first one, you expect a fourth degree equation for a, yet there are apparently only 3 values for a. – coffeemath Oct 10 '12 at 15:26