Solve the simultaneous equations for real numbers $x$ and $y$: $ \sqrt{x+a} + \sqrt{x-a} = 3 $ and $ x+y=5 $

Question

Question: Let $a$ be a real number. Solve the simultaneous equations for real numbers $x$ and $y$:

$$ \sqrt{x+a} + \sqrt{x-a} = 3 $$

$$ x+y=5 $$

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My attempt:

Consider the first equation:

$$ \sqrt{x+a} + \sqrt{x-a} = 3 $$

$$ \Leftrightarrow (\sqrt{x+a} + \sqrt{x-a})^2 = 3^2 $$

$$ \Leftrightarrow x+a+x-a + 2\sqrt{x^2-a^2} = 9 $$

$$\Leftrightarrow 2\sqrt {x^2-a^2} = 9 - 2x $$

$$ \Leftrightarrow 4(x^2-a^2) = 81 - 36x + 4x^2 $$

$$ \Leftrightarrow 4x^2 - 4a^2 = 81-36x + 4x^2 $$

$$ \Leftrightarrow 36x = 4a^2 + 81 $$

$$ \Leftrightarrow x = \frac{1}{36} \left(4a^2+81\right) $$

Subbing this into the second equation

$$ x+y = 5 $$

$$ \Leftrightarrow \frac{1}{36} \left(4a^2+81\right) + y = 5$$

$$ \Leftrightarrow 4a^2 + 81 + 36y = 180 $$

$$ \Leftrightarrow 36y = 99 - 4a^2 $$

$$ \Leftrightarrow y = \frac{11}{4} -\frac{a^2}{9} $$

What should I do now?

Answer 1

We see that $$x = 5-y$$ $$\implies \sqrt{x+a} + \sqrt{x-a} = 3$$ $$\implies \sqrt{5-y+a} + \sqrt{5-y-a} = 3$$ $$\implies (\sqrt{5-y+a} + \sqrt{5-y-a})^2 = 10 + 2\sqrt{5-y+a}\sqrt{5-y-a} - 2y = 9$$ $$\implies 10 + \sqrt{(5-y+a)(5-y-a)} - 2y = 9$$ $$\implies 10 + \sqrt{(y-5)^2-a^2} - 2y = 9$$ $$\implies 2\sqrt{(y-5)^2-a^2} = 2y - 1$$ $$\implies (2\sqrt{(y-5)^2-a^2})^2 = 4((y-5)^2 - a^2) = (2y-1)^2 = 4y^2-4y+1$$ $$\implies 100-4a^2-40y+4y^2 = 4y^2-4y+1$$ $$\implies 100 - 4a^2 - 36y - 1 = 0$$ $$\implies 99 - 4a^2 - 36y = 0$$ $$\implies -36y = -99 + 4a^2 \implies y = \frac{99}{36} - \frac{4}{36}a^2$$ $$\implies y = \frac{11}{4} - \frac{1}{9}a^2.$$ $$\implies x = 5-y = 5-\frac{11}{4}+\frac{1}{9}a^2 = \frac{9}{4}+ \frac{1}{9}a^2.$$

Answer 2

It is assumed without loss of generality that $a\ge 0$.

Let $x = a(\sec^2{\theta}+\tan^2{\theta})$

This is valid since $x \ge a$ and the least value of $\sec^2{\theta}+\tan^2{\theta}$ is $1$

Now, consider the interval $0\le x\le\frac{\pi}{2}$

Then the first equation simplifies to

$$\sec{\theta}+\tan{\theta} = \frac{3}{\sqrt{2a}}$$

This can be tediously bashed using the Weierstrass Substitution $t = \tan{\frac{\theta}{2}}$

Then use that to obtain the corresponding $x$ value, and then $y$.

Similar arguments hold for the remaining three quadrants.

Answer 3

Let us set that $\alpha = \sqrt{x+a}$ and $\beta = \sqrt{x-a}$. Thence we obtain $\alpha + \beta = 3$ and $\alpha^{2} - \beta^{2} = 2a$, which means that \begin{align*} \alpha^{2} - \beta^{2} = (\alpha+\beta)(\alpha-\beta) = 3(\alpha-\beta) = 2a \Leftrightarrow \alpha - \beta = \frac{2a}{3} \end{align*} From this system of equations, we obtain the following solution $\displaystyle(\alpha,\beta) = \left(\frac{3}{2}+\frac{a}{3},\frac{3}{2}-\frac{a}{3}\right)$. Hence:

\begin{align*} \alpha = \sqrt{x+a} & = \frac{3}{2} + \frac{a}{3} \Leftrightarrow x + a = \left(\frac{3}{2} + \frac{a}{3}\right)^{2}\Leftrightarrow x = \left(\frac{3}{2} + \frac{a}{3}\right)^{2} - a \end{align*}

Whenever $x\geq-a$ and $\displaystyle a\geq-\frac{9}{2}$. Analogously, we obtain:

\begin{align*} \beta = \sqrt{x-a} = \frac{3}{2}-\frac{a}{3} \Leftrightarrow x -a = \left(\frac{3}{2}-\frac{a}{3}\right)^{2} \Leftrightarrow x = \left(\frac{3}{2}-\frac{a}{3}\right)^{2} + a \end{align*}

Whenever $x\geq a$ and $\displaystyle a\leq\frac{9}{2}$. Finally, you are able to find the solution set for $x\geq|a|$ and $|a| \leq 4.5$ if you are looking for real solutions. Hope this helps.