# Solutions to $\sqrt{x}+y=6,x^2+y^2=90$

$$\begin{gather} \sqrt{x}+y=6 \tag{1} \\ x^2 + y^2 = 90 \tag{2} \end{gather}$$

WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve for the two variables. I used the method of substitution to finally obtain $x(x+1)=6(21-2y)$. Then I figured that both sides must be non-negative, and hence the value of $y$ has to be less than $11$. Then we test for values less than $11$ which makes $6(21-2y)$ a product of two consecutive numbers. WE get $3$ as a value of $y$, and also $7$. But putting $7$ in (1) does not work.

But there is a flaw in my reasoning. When I get to the 'consecutive' part, I start assuming $x$ and $y$ are integers. Since the person who asked me is not familiar with complex numbers, WE can assume that the solutions are real. However, I am interested in extending the values of $x$ and $y$ beyond the reals, if there are any.

NOTE: From (2), we get the value of $y^2$ in terms of $x$. Then we figure out the value of $x$ in (1) and then substitute $y$ for $90-x^2$, which gives us $x(x+1)$.

• Is it $\sqrt{x}$ or $\sqrt{x+y}$? – Amzoti Sep 8 '13 at 13:20
• @Amzoti,The former.Note that (x,y)=(9,3) is an obvious solution. – rah4927 Sep 8 '13 at 13:21
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• @Arthur Fischer,The edited version is perfect. – rah4927 Sep 8 '13 at 13:34

## 3 Answers

$$\sqrt x + y = 6\tag{1}$$ $$x^2 + y^2 = 90\tag{2}$$

I'll pursue a "different take" and solve for $y$ in terms of $x$:

First, we can consider the task as one of finding the points of intersection of the curve $\sqrt x + y = 6$, and the circle $x^2 + y^2 = 90$. Plotting both equations in Wolfram Alpha gives us: It is quite evident that there is one and only one point of intersection: $(x, y) = (9, 3)$.

Second approach, more algebraic:

From $(1)$, $y = 6-\sqrt x$.

Substituting $y = 6 -\sqrt x$ into $(2)$ gives us

\begin{align} x^2 +(6 - \sqrt x)^2 & = 90 \\ \\ x^2 + 36 - 12\sqrt x + x & = 90 \\ \\ x^2 + x - 12\sqrt x = 54\tag{3}\end{align}

Now, putting $t = \sqrt x$ or $t^2 = x$ gives us the equation:

$$t^4 + t^2 - 12 t - 54 = 0$$

$t = 3$ is one root (and this corresponds to $x = t^2 = 9$. Using polynomial division to factor gives us $$t^4 + t^2 - 12 t - 54 = (t - 3)(t^3 + 3t^2 +10t + 18) = 0$$

Only three other possible roots: one is real, and two complex. The approximate value of the second real root $t$ is $-2.1887$. But this implies $t = \sqrt x \approx -2.1887$ which is impossible, since the square root function returns the positive root. So in the end, our only fruitful solution remains $x = 9,\;y = 3$.

(If you're curious, It's a very elaborate process to find the exact form for the second real solution $t$: From Wolfram, here's a step by step: • @amWhy: Nice details and use of computer to augment! +1 – Amzoti Sep 8 '13 at 14:40
• Thanks, @Amzoti! I would have preferred to link the Wolfram component, but that doesn't work for everyone in terms of accessing it. At any rate, the process is instructive, so I'll leave it stand. – Namaste Sep 8 '13 at 14:42
• @amwhy,Excellent.I like the first approach more than the algebraic one.Is there any practical way to factorize other than knowing that one of the real roots is 3? – rah4927 Sep 8 '13 at 15:34
• the rational root theorem (see Wikipedia) gives us a head start with identifying potential rational roots, based on the first and last coefficients. – Namaste Sep 8 '13 at 15:37

So we have $$\sqrt{x}=6-y$$ which we substitute into the second equation to give $$(6-y)^4+y^2=90.$$ This is equivalent to finding the roots of the polynomial $$f(y):=(6-y)^4+y^2-90.$$ The Fundamental Theorem of Algebra implies this has exactly $4$ (not necessarily distinct) roots.

You mentioned one root, $3$, in your comment. We can find the other roots comptuationally (WolframAlpha); there's closed formulas for them:

$$7+\frac{\sqrt{45+\sqrt{3054}}}{3^{2/3}}-\frac{7}{\sqrt{3(45+\sqrt{3054})}}$$

$$7-\frac{(1+i\sqrt{3})\sqrt{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1-i\sqrt{3})}{2\sqrt{3(45+\sqrt{3054}})}$$

$$7-\frac{(1-i\sqrt{3})\sqrt{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1+i\sqrt{3})}{2\sqrt{3(45+\sqrt{3054}})}$$

Then we substitute them into $\sqrt{x}=6-y$ to get the $x$-values.

The real root listed above implies $\sqrt{x} \approx 6-8.1887$ which has no solution. Thus the solutions are $(9,3)$ and two other complex roots.

• Could you tell me more about the closed formulas? – rah4927 Sep 8 '13 at 15:38
• What else is there to say? The roots of the polynomial could also be found by hand by solving the cubic (since we know $3$ is a root of the polynomial, we take take out the factor of $(y-3)$) and there's a formula for roots of cubics. There's also a formula for the square roots of complex numbers for the later part. – Rebecca J. Stones Sep 8 '13 at 16:22
• Thanks for the link to the formula for roots of cubics. – rah4927 Sep 8 '13 at 16:27

Another way to show there is only one real root...
$\sqrt{x} = 6 - y \implies y \le 6$

So we are looking for roots of $f(y) = (y-6)^4 + y^2 - 90$ where $y \le 6$. We already have $y=3$ as a root from observation.

Using $f(y) = (y-3)(y^3 -21y^2 +154y - 402)$, we need to check if the cubic factor has any real roots in the interval. We note,

If $g(y) = y^3 -21y^2 +154y - 402, g(6) = -18 < 0$ and $g'(y) = 3y^2 - 42y + 154 > 0$ so the cubic has no root with $y < 6$. Hence $(x, y) = (9, 3)$ is the only real solution.