Solving simple quadratic - Wolfram Alpha confusion?

$$(2\sqrt 2 - 2)x^2 + \sqrt8 x + (1+\sqrt 2)=0$$

Now the discriminant of this is $$0$$, so it has one real repeated root. A plot on Desmos confirms this.

However, Wolfram Alpha displays the following (see image). The solution contains $$i$$ and doesn't agree with what it should be $$= -1.707...$$

What is happening?

[Solution should be as I said above because $$x = \frac{-\sqrt8 \pm 0 }{2\sqrt2 - 2} = -1 - \frac{1}{2} \sqrt 2 \approx -1.707...$$ after simplifying]

• What image? I don't see one. – Cameron Williams Mar 12 at 17:43
• Those are different expressions. One involves $2\sqrt 2 - 2$ which is positive and the other involves $\sqrt 2 - 2$ which is negative. So they'll have different solutions. – fleablood Mar 12 at 17:46
• No sorry, I've changed it now!! – PhysicsMathsLove Mar 12 at 17:46
• Wolfram Alpha tends to do numerical solutions with certain kinds of polynomials and if you look at the imaginary part, it's $10^{-8}$, so it's doing a numerical solution and getting a really small imaginary part. For all intents and purposes, consider it to be pure real. – Cameron Williams Mar 12 at 17:46
• A plot on Desmos confirms this --- To me this is like checking that the city has picked up your garbage by looking to see that your garbage can at the end of your driveway is empty, and then confirming that your garbage can must be empty by going online to see if today was a garbage pickup day. – Dave L. Renfro Mar 12 at 18:08

That $$\pm4.21468\times10^{-8}i$$ results from a rounding error and should be seen as $$0$$. So, the numerical answer is actually $$.603553\times(-2.82843)$$, which is indeed about $$-1.707$$.