Question

Please help me with Algebra 2 :) will give both fan and medal.

Answer 1

can u copy and paste the qustion

Answer 2

not the answers

Answer 3

Sure, the question says
2x²+x+2=0

Answer 4

\(\color{#000000 }{ x=\displaystyle \frac{ -1\pm\sqrt{1^2-4\cdot 2 \cdot 2 \color{white}{|}} }{2\cdot 2} }\) (Via the quadratic formula)

Answer 5

simplify, everything you could,
and be aware that \(\sqrt{-b}=i\sqrt{b\color{white}{~|}}\)

Answer 6

That's the question and answers @alivejeremy

Answer 7

Okay, thank you :) I'll try to work it out. Thank you very much! @SolomonZelman

Answer 8

Would it be A or B? I just tried working part of it

Answer 9

I think it's A.

Answer 10

the square root has \(\pm\), and if you want to know why I can post an explanation:)

Answer 11

(In this case does, but not always)

Answer 12

Yes please :)

Answer 13

Suppose you are solving the equation
\(\color{#000000 }{ \displaystyle x^2-1=8 }\)
When you add 1 to both sides you get,
\(\color{#000000 }{ \displaystyle x^2=9 }\)
And then you take the square root of both sides (right?)

Most people think that kaboom, magic, and ...
\(\color{#000000 }{ \displaystyle x=\pm\sqrt{9}=\pm 3}\)
HOWEVER, THAT HAS TO DO WITH THE DEFINITION OF THE ABSOLUTE VALUE!!

So, let's go back a step to taking the square root of both sides,
\(\color{#000000 }{ \displaystyle \sqrt{x^2}=\sqrt{~9~} }\)

And here comes in the definition of the abs. value;
Absolute value of (let's call it \(z\)) is defined as;
\(\color{#000000 }{ \displaystyle |z|=\sqrt{(z)^2} }\)
(When you plug in -z or +z into the \(\sqrt{(z)^2}\), you get z in either case...)

So what happens when taking the square rooot of both sides is,
\(\color{#000000 }{ \displaystyle \sqrt{x^2}=\sqrt{~9~} }\)
\(\color{#000000 }{ \displaystyle |x|=3 }\)
(Simplifying the right side because we know √9=3, and on the left I applied the definition of the absolute value),

And from there we get
\(\color{#000000 }{ \displaystyle x=\pm3 }\)

Answer 14

If I had something like
\(\color{#000000 }{ \displaystyle x^3=8 }\)
\(\color{#000000 }{ \displaystyle \sqrt[3]{x^3}=\sqrt[3]{8}}\)
\(\color{#000000 }{ \displaystyle x=2}\)
(-2 is not an option, because when the powers are even there is no absolute value, since when we plug in a and -a into \(\sqrt[3]{(z)^3}\), we get different results.)

Answer 15

So, when solving your equation, ...

Answer 16

Prove for quadratic formula, through completing the square

\(\color{black}{ \displaystyle \color{#009900}{\rm a}x^2+\color{#009900}{\rm b}x= -\color{#009900}{\rm c} }\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x\right)= -\color{#009900}{\rm c} }\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\color{black}{-}\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\right)= -\color{#009900}{\rm c}}\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a^2}}\right)= -\color{#009900}{\rm c}}\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm a\frac{b^2}{4a^2}}= -\color{#009900}{\rm c}}\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a}}= -\color{#009900}{\rm c}}\)

\(\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm c}+\color{#009900}{\rm \frac{b^2}{4a}}}\)

\(\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}\)

\(\color{black}{ \displaystyle \sqrt{\left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2}= \sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\LARGE \color{red}{\Longleftarrow}\)

\(\color{black}{ \displaystyle \left|x+\color{#009900}{\rm \frac{b}{2a}}\right|= \sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\LARGE \color{red}{\Longleftarrow}\)

\(\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)= \pm \sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\LARGE \color{red}{\Longleftarrow}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{4ac}{4a^2}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{4a^2}}}}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{(2a)^2}}}}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{\sqrt{(2a)^2}}}}\)

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{|2a|}}}\)

and |2a| makes no difference, because we already have \(\pm\) by the square root, so

\(\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{2a}}}\)

Answer 17

In any case, in your problem,

\(\color{#000000 }{ x=\displaystyle \frac{ -1\pm\sqrt{1^2-4\cdot 2 \cdot 2 \color{white}{|}} }{2\cdot 2} }\) (Via the quadratic formula)

\(\color{#000000 }{ x=\displaystyle \frac{ -1\pm\sqrt{-15 \color{white}{|}} }{4} }\)

\(\color{#000000 }{ x=\displaystyle -\frac{ 1\pm\sqrt{-15 \color{white}{|}} }{4} }\)

Answer 18

then take out the i from there

Answer 19

Thank you so so so much. This is very helpful!

Answer 20

yw...

Answer 21

(I don't want to overtype it) good luck!

Answer 22

No, this is perfect. Thank you!

Answer 23

:)