In MAO 1991,
Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$
My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is very tedious. Thanks in advance.
$\endgroup$3 Answers
$\begingroup$We can construct some nice solutions after having known the answer a posteriori.
Put $x=-y$ where $y>0$ (as the quantity under square root must be positive). Hence
$$\sqrt{40+9y}-2\sqrt{7+y}=\sqrt{y}$$ from which it can be seen that $y=9$ (i.e. $x=-9$) satisfies the equation.
Alternatively, put $u=x+9$ (i.e. $x=u-9$) in the original equation, giving $$\begin{align} \sqrt{40-9(u-9)}-2\sqrt{7-(u-9)}&=\sqrt{9-u}\\ \sqrt{121-u}-2\sqrt{16-u}&=\sqrt{9-u}\end{align}$$ Since $121, 16, 9$ are perfect squares, we try putting $u=0$ to make the square roots disappear and in doing this we find that the equation is satisfied. Hence the solution is $u=0$ i.e. $x=-9$.
$\endgroup$ 4 $\begingroup$Substitution $3.5-x=t$ seems to help because $-x$ and $7-x$ are symmetric around the point $3.5$ you get $$\sqrt{9t+8.5}=\sqrt{t-3.5}+2\sqrt{t+3.5}\\9t+8.5=t-3.5+4t+14+4\sqrt{t^2-3.5^2}\\4t-2=4\sqrt{t^2-3.5^2}\\(2t-1)^2=4(t^2-3.5^2)\\4t^2-4t+1=4t^2-4\cdot 3.5^2\\4t=4\cdot 3.5^2+1\\4t=(2\cdot 3.5)^2+1\\4t=50\\t=12.5\\x=-9$$
$\endgroup$ 1 $\begingroup$An $x$ that solves your equation also solves the system \begin{align} 40&&-9x&-a^2&=0\\ 7&&-x&-b^2&=0\\ &&-x&-c^2&=0\\ &&a-2b&-c&=0\\ \end{align} If we subtract the second and the third equation we get $$-b^2+c^2+7=0.$$ It turns out that $$-9(-b^2+c^2+7)=-(3b-4c)(-3b-4c)-(-7c^2+63).$$ Therefor one solution is $c^2-9=0$ and $3b-4c=0$. Both solutions to $c$ work. If we pick $c=3$ we get $x=-9$ from the third equation and $b=4$ which makes $x=-9$ consistent with the second equation.
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