Prove that, there are 4 real roots of system of equations: $\begin{cases} y^2+x=11 \\ x^2+y=7 \end{cases}$

Question

How can I prove that, there are 4 real roots of this system of equation?

Solve for real numbers:

$$\begin{cases} y^2+x=11 \\ x^2+y=7 \end{cases}$$

My attempts:

$$(7-x^2)^2+x=11 \Longrightarrow x^4 - 14 x^2 + x + 38=0 \Longrightarrow (x - 2) (x^3 + 2 x^2 - 10 x - 19) = 0$$

So, we have $x=2, y=3.$

Now , how can I prove that all other roots are also real? Becasue, Wolfy says, there are $4$ real roots. To do this, there is probably no escape from the derivative. Do I think right?

Answer 1

well, as I said, adding the two equations gives a new equation that defines a circle, center at $\left( \frac{-1}{2}, \frac{-1}{2} \right)$

This gives one way to find numerical solutions, name $$ x = \frac{-1 + \sqrt{74} \cos t}{2} \; \; , \; \; \; y = \frac{-1 + \sqrt{74} \sin t}{2} $$ and solve either parabola numerically for $t.$

The value you already know is $$ t = \arctan \frac{7}{5} \approx 0.950546841 $$

For $ \frac{\pi}{2} < t < \pi, $ my calculator says $t \approx 1.889602434,$ then $x \approx -1.848126529$ and $y \approx 3.584428340.$

For $ \pi < t < \frac{\pi}{2} , $ my calculator says $t \approx 4.008643846,$ then $x \approx -3.283185989$ and $y \approx -3.779310256.$

For $ \frac{3\pi}{2} < t < 2 \pi, $ my calculator says $t \approx 5.717577494,$ then $x \approx 3.131312516$ and $y \approx -2.805118090.$

It really is worth practicing drawing

Answer 2

The second equation implies $y=7-x^2$ which, when plugged into the first equation, renders

$(7-x^2)^2+x=11$

$x^4-14x^2+x+38=0$

If $|x|$ is as large as $19$, then $x^4$ dominates the other terms of the polynomial so the only rational roots worth further consideration are $\pm1, \pm 2$. Of these $x=2$ holds giving the factorization

$(x-2)(x^3+2x^2-10x-19)=0$

The cubic factor is negative at $x=0$ and as $x\to-\infty$, but positive ($+1$) at $x=-2$, so a pair of negative roots is assured for $x$; and Descartes' Rule of Signs assures a positive root. These together with the previously found rational root $x=2$ constitute four real roots for $x$, and thence a real value for $y=7-x^2$ associated with each of these.

Answer 3

You wonder about the possible zero's of function$$f(x)=x^3 + 2 x^2 - 10 x - 19$$ Consider $$f'(x)=3x^2+4x-10 \qquad \text{and} \qquad f''(x)=6x+4$$ The firs derivative cancels at $$x_1=-\frac{1}{3} \left(\sqrt{34}+2\right)\qquad \text{and} \qquad x_2=\frac{1}{3} \left(\sqrt{34}-2\right)$$ $f''(x_1)=-2 \sqrt{34}<0$ shows that $x_1$ corresponds to a maximum and $f''(x_2)=2 \sqrt{34}>0$ shows that $x_2$ corresponds to a minimum.

Now $$f(x_1)= \frac{1}{27} \left(68 \sqrt{34}-317\right) >0 \qquad \text{and} \qquad f(x_2)=-\frac{1}{27} \left(68 \sqrt{34}+317\right) <0$$

So, three real roots for the cubic.

If you apply the trigonometric method for cubic equations, the roots are given by $$x_k=\frac{2}{3} \left(\sqrt{34} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{317}{68 \sqrt{34}}\right)\right)\right)-1\right)\qquad \text{for} \qquad k=0,1,2$$

Answer 4

Note that the cubic equation $x^3+bx^2+cx+d=0$ has three real solutions if its discriminate $\Delta$ satisfies

$$\Delta = 18bcd-4b^3d+b^2c^2-4c^3-27d^2>0$$

So, for the equation,

$$(x-2)(x^3 + 2 x^2 - 10 x - 19)=0$$

the discriminate of its cubic factor is

$$\Delta = 2101 > 0$$

Thus, it has four real roots.