Trying to understand roots of quadratic equations

[Martyn Arthur]

TL;DR

Finding a solution +- to the quadratic formula

I understand the basic maths but I am getting varying answers as to whether these are real distinct roots or not. Could you please explain the mechanism for deciding this. Thanks in anticipation.

 

[Mark44]

TL;DR Summary: Finding a solution +- to the quadratic formula

I understand the basic maths but I am getting varying answers as to whether these are real distinct roots or not. Could you please explain the mechanism for deciding this. Thanks in anticipation.View attachment 351948

What does your post have to do with complex numbers?
Edit: Thread title originally mentioned complex numbers, but has been corrected.

The two solutions you showed are real numbers. Technically, all real numbers also are complex, but with a coefficient of zero for the imaginary number i. You can verify for yourself that they are distinct by using a calculator to get approximate values for each.

If there are people who are telling you that $-1 + 2\sqrt 3$ and $-1 - 2\sqrt 3$ aren't distinct, tune them out as they don't know what they're talking about.

OTOH, if you question is about whether $\frac{-2 \pm 4\sqrt 3}2$ represent different numbers than the two in the preceding paragraph, they don't.

Can you be more specific in what you're asking?

 

[Mark44]

Incidentally, the two roots you show are solutions of the quadratic equation $x^2 + 2x - 11 = 0$.

Also, one doesn't find a solution to the quadratic formula -- one uses the quadratic formula to find solutions of a quadratic equation.

 

[fresh_42]

If you want to understand the notation of $\pm$, and this is only a notational shortcut, you have to look at it from the start. We start with $r^2+2r-11=0.$ Next, we are completing the square as it is called. $r^2+2r$ reminds of the binomial theorem $r^2+2r+1=(r+1)^2$ so we bring our equation into such a pattern: $0=r^2+2r+1-12=(r+1)^2-12,$ means $12=(r+1)^2.$ This equation has two solutions: $r+1=\sqrt{12}$ and $r+1=-\sqrt{12}.$ Now, the notation kicks in. We write for convenience $r+1=\pm \sqrt{12}$ to cover both cases in one formula. However, it means that we calculate with two separate equations at the same time from now on.
\begin{align*}
r+1&=\pm \sqrt{12}=\pm 2\sqrt{3} \Longleftrightarrow r=-1\pm 2\sqrt{3}
\end{align*}
It literally means $r^2+2r-11=0 \Longleftrightarrow r\in \left\{-1+2\sqrt{3}\, , \,-1-2\sqrt{3}\right\}.$

 

[Martyn Arthur]

So naming the 2 solutions x we have +x and -x?

 

[fresh_42]

So naming the 2 solutions x we have +x and -x?

No. $-1-2\sqrt{3}\neq -(-1+2\sqrt{3}).$ In fact, we have so-called conjugates. $-2\sqrt{3}=-(+2\sqrt{3})$ but the constant term $-1$ is in both solutions the same. If the roots were complex numbers, say
$$
0=x^2-2ax+(a^2+b^2)=(x-(a+\boldsymbol i b))\cdot (x-(a-\boldsymbol i b))
$$
then $a+\boldsymbol i b$ and $a-\boldsymbol i b$ are complex conjugate numbers. The principle is the same whether it is $\boldsymbol i$ or $2\sqrt{3}.$ It means geometrically that $f(x)=a_2x^2+a_1x+a_0$ is a parabola and the roots are equidistant from the symmetry axis in the middle. This axis can be the y-axis ($x=0$) but does not have to be. In your example, it was the straight $x=-1.$

 

[Martyn Arthur]

Thank you; I need to go away and think about it.
Martyn

 

[Mark44]

So naming the 2 solutions x we have +x and -x?

To elaborate on what was already said, the two solutions are two different values of x. Also, if $x = -1 + 2\sqrt 3$, then -x would be $+1 - 2\sqrt 3$. The latter is not a solution to the quadratic equation you're asking about. The two solutions of the quadratic equation are not negatives of each other: they are conjugates.

 

[Vanadium 50]

We probably should close (or retitle) this thread.

 

[pinball1970]

Thank you; I need to go away and think about it.
Martyn

https://en.wikipedia.org/wiki/Complex_number