Find the roots of equation based on some geometry hints

Question

Plots of the equations $y = 8 - x^2$ and $|y|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$. We have to solve the equation $$8-x^2=\sqrt{8+x}$$

Answer 1

If $y=8-x^2$ and $|y|=\sqrt{8+x}$, then $y^2-y=(8+x)-(8-x^2)=x+x^2,$

so $y(y-1)+x(-1-x)=0,$ so $y(y-1-x)+x(y-1-x),$ so $(y+x)(y-1-x)=0,$

i.e., $y=-x$ or $y=1+x$. Therefore $x$ must be a solution of $8-x^2=-x$ or $8-x^2=1+x$.

Can you solve these quadratic equations? Note that you want solutions where $y=8-x^2\ge0$.

Answer 2

Hint:

We intersect the parabolas

$$y=8-x^2$$ and $$y^2=x+8.$$

Let us form a pencil of conics by the same four intersection points and find the one that degenerates in a pair of lines.

$$y+x^2-8+\lambda(y^2-x-8)=0$$ has a double point where (by canceling the gradient)

$$2x-\lambda=0,\\1+2\lambda y=0.$$

We draw $x,y$ and plug in the equation of the pencil,

$$-\frac1{2\lambda}+\frac{\lambda^2}4-8+\lambda\left(\frac1{4\lambda^2}-\frac\lambda2-8\right)=0$$ or

$$\lambda^3+32\lambda^2+32\lambda+1=0.$$

By inspection, $\lambda=-1$ is a root, giving the degenerate conic

$$(x+y)(x-y)+(x+y)=0.$$

The rest is straightforward (two quadratic equations; ensure $y\ge0$).

Answer 3

$y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ are symmetric w.r.t. the line $y=-x$ implies the tangent lines at the intersection points are reflected over $y=-x$.

Refer to the graph:

$\hspace{3cm}$

The line $y_1=ax+b$ is reflected over $y=-x$ to the line $y_2=\frac1ax+\frac ba$. It implies $a\cdot \frac1a=1$.

Reference: $y_1=8-x^2$ and $|y_2|=\sqrt{8+x}$ (or $y_2=\pm \sqrt{8+x}$).

Finding $x_2$ (blue lines): $$y_1'(x_2)\cdot y_2'(x_2)=1 \Rightarrow -2x_2\cdot \frac1{2\sqrt{8+x_2}}=1 \Rightarrow x_2=\frac{1-\sqrt{33}}{2}.$$

Finding $x_4$ (purple lines): $$y_1'(x_4)\cdot y_2'(x_4)=1 \Rightarrow -2x_4\cdot \left(-\frac1{2\sqrt{8+x_4}}\right)=1 \Rightarrow x_4=\frac{1+\sqrt{33}}{2}.$$

Finding $x_1$ and $x_3$ (red and green lines): $$\begin{cases}y_1'(x_1)\cdot y_2'(x_3)=1\\y_1'(x_3)\cdot y_2'(x_1)=1 \end{cases} \Rightarrow \begin{cases}-2x_1\cdot \frac1{2\sqrt{8+x_3}}=1 \\ -2x_3\cdot \left(-\frac1{2\sqrt{8+x_1}}\right)=1\end{cases} \Rightarrow \begin{cases}x_1^2=8+x_3\\ x_3^2=8+x_1\end{cases} \Rightarrow \\ (x_1-x_3)(x_1+x_3)=x_3-x_1 \Rightarrow x_1=-1-x_3 \Rightarrow x_3^2+x_3-7=0 \Rightarrow \\ x_3=\frac{-1+\sqrt{29}}{2} \Rightarrow x_1=\frac{-1-\sqrt{29}}{2}.$$

Answer 4

This issue necessitates a graphical representation

Fig. 1 : Curves with equations $y=8-x^2$ (cyan and magenta), $y=\sqrt{x+8}$ (red) and $y=-\sqrt{x+8}$ (green). We have figured as well line with equation $y=x+1$ (see the solution by @J. W. Tanner).

Initial equation $8-x^2=\color{red}{+}\sqrt{8+x}$ and its "twin equation" $8-x^2=\color{red}{-}\sqrt{8+x}$ are equivalent to :

$$(8-x^2)^2-(8+x)=0$$

Luckily, this 4th degree equation has the following factorization :

$$(x^2 + x - 7)(x^2 - x - 8)=0$$

Therefore, it remains to compute the roots of the two quadratic factors.

$$x_1=\dfrac12(-1-\sqrt{29}), \ \ x_2=\dfrac12(-1+\sqrt{29}), \ \ x_3=\dfrac12(1-\sqrt{33}) \ \text{and} \ x_4=\dfrac12(1+\sqrt{33}).$$

Clearly, the solutions of the initial equation are the least extremes $x_2$ and $x_3$, whereas the other "spurious" ones $x_1$ and $x_4$ are roots of the twin equation.

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Now, let us find back "geometrically" the elegant solution given by @J.W. Tanner, involving the use of straight lines $y=-x$ (no question...) and $y=x+1$ (less intuitive...).

If you happen to know pencils of conic sections, consider the pencil (family of conic curves) given by linear combinations of the two parabolas' equations :

$$\alpha(y^2-8-x)+\beta(y-8+x^2)$$

It is well known that an equivalent description in such a case is the set of curves passing by the four common points of the parabolas defining the pencil.

Taking $\alpha = 1$ and $\beta=-1$ gives

$$y^2-x^2-x-y=0 \ \iff \ (y+x)(y-x-1)=0 \ \iff \ \begin{cases}y&=&-x\\y&=&x+1\end{cases}$$

explaining that the union of straight lines with equations $y=-x$ and $y=x+1$ (a degenerated conic curve) belongs to the pencil and as such, passes through the four intersection points of the two parabolas.

Answer 5

The given two equations are one and the same. They are swapped in $(x,y)$ with sign change.

I am sorry that I am not able to understand why a given repeated procedure should unconditionally be useful in solution finding.

Is it a text-book problem? What is the topic about?

An indirect claim that any polynomial (or by implication any transcendental equation) can be solved by means of such anti-symmetric inverse function logic as means for a solution may be difficult to justify, or so it seems to me. $x+y=0$ etc... procedure maybe just associated with such operation performed.

The geometrical hint is just an observation not an indispensable input premise for solution,imho.

We have by the straight procedure indicated:

$$ 8-x^2= y= -x\quad \rightarrow x^2-x -8 = 0, $$

$$ x_1=(1-\sqrt{33})/2 ;\, x_2 = (1+\sqrt{33})/2$$

The procedure adopted cannot be exhaustive. One needs to proceed with a conventional solution. Resorting to squaring to remove radical will introduce extraneous solutions to be later removed.

$$ y= 8-x^2; \quad y^2 = x^4-16x^2+64 \tag1$$

$$ y^2=8+x \tag2 $$

Equate $y^2$ in 1) and 2)

$$ x^4-16 x^2-x + 56=0 \tag3$$

Two roots are found already and in order to find the other two factors/roots in the fourth order polynomial we can write:

$$x^4-16 x^2-x + 56 = (x^2-x-8)((x^2+\lambda x -7)$$

Evaluating undetermined coefficients after expansion we find three $\lambda$ values:

$$ \lambda = (1,-15,8) \tag 4$$

We take first value, eliminating next two $ \lambda =(-15,8)$ as spurious/extraneous roots confirmed in a graphical method later ( not fully convincing though ) getting expressions for $y$

$$ y=(x^2-x-8) (x^2+x-7)=0 \tag 5 $$ $$ y_2=(x^2-x-8) (x^2-15 x-7)=0 \tag {5a} $$ $$ y_3=(x^2-x-8) (x^2+ 8x-7)=0 \tag {5b} $$

The other two real roots of second quadratic are $$ x_3= (-1+\sqrt{29})/2,\, x_4=(-1-\sqrt{29})/2 \tag6 $$

(5) is plotted (scaled down factor 100 on y axis) with all four real roots shown.

Please note that (5a) and (5b) are all valid solutions because there are four real valid roots in the all three sets which are calculated as above and found to be correct as per expectation on extended plot domains (not graphed here).