Question

Check my answer?
Solve. 2x2 – 8x – 16 = 0

x = 2 ± 2 sqrt 3

x = 4 ± 2 sqrt 5

x = 2 ± 4 sqrt 2

x = 4 ± 2 sqrt 3

I think its the thrid one!

Answer 1

*third

Answer 2

unfortunately no, and i bet i know exactly the mistake you made, it is easy to do

Answer 3

i bet you divided by 2 in the numerator right?

Answer 4

lol yea i did.

Answer 5

unfortunately you cannot divide inside the radical by 2, you have to write in simplest radical form first
lets do it the easy way

Answer 6

you can check these on you graphing calculator if you have one

Answer 7

Hint: Factor by grouping

Answer 8

@Outkast3r09 i dont have one:/

Answer 9

@Hero the answer is a radican

Answer 10

quadratic formula would also work

Answer 11

first divide everything by 2 to get
\[x^2-4x-8=0\]
the we can use the quadratic formula or what is easier is to complete the square
\[x^2-4x-8=0\]
\[x^2-4x=8\]
\[(x-2)^2=8+4=12\]
\[x-2=\pm\sqrt{12}=\pm2\sqrt{3}\]
\[x=2\pm2\sqrt{3}\]

Answer 12

No... no factoring... QUADRATIC FORMULA.

Answer 13

now lets use the formula and see what happens to get the same answer
the formula is harder because it forces a denominator on you, but there is none

Answer 14

okay thanks

Answer 15

either method would work, you would have to simplify at the end

Answer 16

\[x^2-4x-8=0\]
\[a=1,b=-4,c=-8\]
\[x=\frac{4\pm\sqrt{(-4)^2-4\times 1\times -8}}{2}\]
\[x=\frac{4\pm\sqrt{48}}{2}\] now you cannot divide 48 by 2 to get \(\sqrt{24}\) because it is not \(48\) is is \(\sqrt{48}\) so you have to write it in simplest radical form before you can divide

Answer 17

@outkast3r09, that did not stop me from factoring by grouping

Answer 18

\[\sqrt{48}=4\sqrt{3}\]so now you have
\[\frac{x=4\pm4\sqrt{3}}{2}\]and NOW you can divide both terms by 2 to get
\[x=2\pm2\sqrt{3}\]