Question

How would you solve this using the quadratice equation? 2x^2-12x+13=0

Answer 1

Quadratic

Answer 2

\[\Large\bf\sf \color{orangered}{a}x^2+\color{royalblue}{b}x+\color{purple}{c}\quad=\quad 0\]And we have,
\[\Large\bf\sf \color{orangered}{2}x^2+\color{royalblue}{(-12)}x+\color{purple}{13}\quad=\quad 0\]

Answer 3

We'll plug this into,\[\Large\bf\sf x\quad=\quad \frac{\color{royalblue}{b}\pm\sqrt{\color{royalblue}{b}^2-4\color{orangered}{a}\color{purple}{c}}}{2\color{orangered}{a}}\]

Answer 4

\[\Large\bf\sf x\quad=\quad \frac{\color{royalblue}{-12}\pm\sqrt{\color{royalblue}{(-12)}^2-4\cdot\color{orangered}{2}\cdot\color{purple}{13}}}{2\cdot\color{orangered}{2}}\]Does it make sense the way I plugged those in?

Answer 5

Do you understand why the 12 is negative?

Answer 6

I've done the problem but i keep getting a different answer than my teacher had, and i plugged it in the right way.

Answer 7

When you squared the 12, did you also square the negative in front of it?
It should become positive.

Answer 8

Yes.

Answer 9

Maybe your teacher's wrong?

Answer 10

So I guess we get...\[\Large\bf\sf x\quad=\quad \frac{-12\pm\sqrt{144-104}}{4}\]Simplifying a little further,\[\Large\bf\sf x\quad=\quad \frac{-12\pm\sqrt{40}}{4}\]Which we can simplify even further.
Does your work much up so far?

Answer 11

match up*

Answer 12

I Have That, Just I Have A Positive 12.

Answer 13

Oh yes, my bad.

Answer 14

\[\Large\bf\sf x\quad=\quad \frac{12\pm\sqrt{40}}{4}\quad=\quad \frac{12\pm2\sqrt{10}}{4}\quad=\quad 3\pm\frac{\sqrt{10}}{2}\]

Answer 15

So you can simplify it pretty far if you want.
Does your teachers result look like that maybe?

Answer 16

No, she has 1/2 (6+ or -squareroot 10)

Answer 17

\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

\[x=\frac{ -(-12)\pm \sqrt{(-12)^{2}-4(2)(13)} }{ 2(2) }\]

\[x=\frac{ 12\pm \sqrt{144-104} }{ 4 }\]

\[x=\frac{ 12\pm \sqrt{40} }{ 4 }\]

\[x=\frac{ 12\pm2\sqrt{10} }{ 4 }\]

Idk, this is how far I got

Answer 18

That's What I Got Aswell ^

Answer 19

\[\Large\bf\sf 3\pm\frac{\sqrt{10}}{2}\]Factoring 1/2 out of each term gives,\[\Large\bf\sf =\frac{1}{2}\left(6\pm \sqrt{10}\right)\]

Answer 20

It's the same result, just written a little differently.

Answer 21

Oh, Okay! I was just confused how she got that.

Answer 22

Do either of you know how to complete squares? I have the answer i believe is correct i'm just not positive.