Question

Solve for x: 2x2 - 4x - 14 = 0....(completing the square) I have the answer up to 2(x-1)^2√7 plzz tell me if I'm right and if I'm not plz explan.

Answer 1

Probably, it's not right...
2x^2 - 4x - 14 = 0
2(x^2 - 2x - 7) = 0
x^2 - 2x - 7 = 0
x^2 -2x +(1-1) -7 = 0
(x^2 -2x+1) -1 -7 = 0
(x-1)^2 -8 = 0
(x-1)^2 = 8
Can you continue?

Answer 2

yes, I think so, thank u very much :D

Answer 3

You're welcome :)

Answer 4

so for the last part I put them both under √'s?.

Answer 5

That is the second last part.

Answer 6

okay, I'm kinda stuck..what is the first to the last part.

Answer 7

Take square root on both sides.

Answer 8

I already said that and u said that waz the 2nd to the last part?

Answer 9

What do you get for that step?

Answer 10

I get x-1 = + or _ √8

Answer 11

than you solve for both sides right?

Answer 12

Yes, you need to isolate x on the left..

Answer 13

for my answer I get x=-8 and x=8?

Answer 14

No!

Answer 15

x=7 and x=-9?

Answer 16

No!
\[x-1 = \pm \sqrt8\]
Add 1 to both sides

Answer 17

soo -1 +8?

Answer 18

omg , I know this is so simply but I'm just not understanding it srry.

Answer 19

No..... you cannot remove the square root. Moreover, you should add 1, instead of minus 1

Answer 20

ohh , so x=1!!

Answer 21

NO!!!!!!

Answer 22

just tell me the answer and then explan it to mee! ....

Answer 23

\[x-1=\pm \sqrt{8}\]Add 1 to both sides\[(x-1)+1=(\pm \sqrt{8})+1\]\[x=1 \pm \sqrt{8}\]

Answer 24

soo 1-2√2?

Answer 25

no it's x - 1 = ±2√2? say somthing....!

Answer 26

heyy person viewing this do you know the answer?

Answer 27

\[2x^{2} -4x-14=0\]
\[2\left(\left(x-\frac{2}{2}\right)^{2}-\left(\frac{2}{2}\right)^{2}-7\right) =0\]
\[2\left(\left(x-1\right)^{2}-(1)-7\right) = 0 \]
\[2\left(\left(x-1\right)^{2}-8\right) =0 \]
\[2\left(x-1\right)^{2} - 16 = 0 \]
\[2\left(x-1\right)^{2} =16 \]
\[\left(x-1\right)^{2} = 8 => \left(x-1\right) = \pm2\sqrt{2} => => x = 1\pm2\sqrt{2} \]

Answer 28

ty :)