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# Ersetzen Sie das Symbol * durch die richtige Zahl,

## Tags: Ersetzen Sie das Symbol * durch die richtige Zahl, um die richtige Gleichung zu erhalten.     RASTA 23:21 Uhr, 26.11.2021

HILFE! Bitte!
Ersetzen Sie das Symbol $\cdot$ durch die richtige Zahl, um die richtige Gleichung zu erhalten.

${\left(2x\phantom{\rule{0.12em}{0ex}}+\cdot \right)}^{2}=4{x\phantom{\rule{0.12em}{0ex}}}^{2}+$ 12xy $+9y\phantom{\rule{0.12em}{0ex}}$

Vielen Dank im Voraus

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."  23:24 Uhr, 26.11.2021

As I told you: $3\cdot y\phantom{\rule{0.12em}{0ex}}$

( And you surely meant $9\cdot {y\phantom{\rule{0.12em}{0ex}}}^{2}$ and not $9\cdot y\phantom{\rule{0.12em}{0ex}}\right)$  RASTA 01:12 Uhr, 27.11.2021

Yes, thank you!
You are right.
${\left(2x\phantom{\rule{0.12em}{0ex}}+\cdot \right)}^{2}=4{x\phantom{\rule{0.12em}{0ex}}}^{2}+$ 12xy $+9{y\phantom{\rule{0.12em}{0ex}}}^{2}$  RASTA 01:14 Uhr, 27.11.2021

The symbol is *.
It looks like a multiplication symbol.  RASTA 01:14 Uhr, 27.11.2021

The symbol is *.
It looks like a multiplication symbol. You were already told the solution $3y\phantom{\rule{0.12em}{0ex}}$ twice (here and in your other thread).
Didn't you notice it or is there anything else unclear?
It is an application of the binomial formula ${\left(a\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}\right)}^{2}={a\phantom{\rule{0.12em}{0ex}}}^{2}+2\cdot a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}+{b\phantom{\rule{0.12em}{0ex}}}^{2}$.
Set $a\phantom{\rule{0.12em}{0ex}}=2x\phantom{\rule{0.12em}{0ex}}$ and you get
${\left(2x\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}\right)}^{2}=4{x\phantom{\rule{0.12em}{0ex}}}^{2}+2\cdot 2x\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}+{b\phantom{\rule{0.12em}{0ex}}}^{2}$
and now compare with
$4{x\phantom{\rule{0.12em}{0ex}}}^{2}+12x\phantom{\rule{0.12em}{0ex}}y\phantom{\rule{0.12em}{0ex}}+9{y\phantom{\rule{0.12em}{0ex}}}^{2}=4{x\phantom{\rule{0.12em}{0ex}}}^{2}+2\cdot 2x\phantom{\rule{0.12em}{0ex}}\cdot 3y\phantom{\rule{0.12em}{0ex}}+{\left(3y\phantom{\rule{0.12em}{0ex}}\right)}^{2}$
and guess what $b\phantom{\rule{0.12em}{0ex}}$ could be. :-)

BTW, if you'd like to see the $\star$ character, simply type two of them. If you type just one, its converted to the small multiplication dot.  RASTA 17:12 Uhr, 27.11.2021

Thnk you so much for posting the original formula. Now I see it clealy. I had some issues at first posting. That is why I repeated my post and maybe my question. But now it is all good.