Solve $1-x^2=5-2x$ graphically

Question

Solve the following equation graphically: $$1-x^2=5-2x$$

To solve the equation graphically, we must draw the graph for each side, member of the equation, and see where they cross, are equal. The $x$ values of these points are the solutions to the equation. $$y_1=1-x^2$$ and $$y_2=5-2x$$

We see that the graphs don't cross; therefore, we don't have solutions to the given equations. According to WolframAlpha, this equation has complex roots. We can't see them in the Cartesian coordinate system, right?

Answer 1

We can make the complex roots visible as real roots if we reflect the graph at the apex of the parabola. Here is a nice answer which explains the procedure. I´m going to show how it works graphically in your case

We can transform the equation to $f(x)=0$

$$f(x)=x^2+bx+c=x^2-2x+4=0$$

Now we can transform the function f to get the corresponding real roots.

$$h(x)=-x^2-bx-\frac{b^2}2+c=-x^2+2x-2+4$$

The graphs looks like below.

The real roots of $h(x)$ are $x_1=1-\sqrt{3}, x_2=1+\sqrt{3}$

Therefore the complex roots of $f(x)$ are $x_1=1-i\cdot \sqrt{3}, x_2=1+i\cdot \sqrt{3}$

Answer 2

The equation is the same as $$x^2-2x+4=0$$ which is solved by the quadratic formula, $$ x=\frac {2\pm \sqrt {4-16}}{2}=1\pm i\sqrt 3 $$

Therefore the roots are complex numbers and we do not see them on the plane $\mathbb {R^2}$ where coordinates are real numbers.

Similar situation happened with polynomials such as $$x^2+1$$ and $$x^4+1$$ We simply can not see the complex roots on the real plane.

Answer 3

When you plotted your curves, you assumed that the $ x $ coordinate is real .

But your second degree equation has no real root. its discriminant is negative.

If you want to solve it graphically, you should plot two surfaces :

$$f(z)=1-z^2$$ and $$g(z)=5-2z.$$