Question

Solve 2x2 + 12x − 14 = 0 by completing the square.

2(x + 6)2 = 26; x = −6 + square root 13, x = −6 − square root 13
2(x + 6)2 = 20; x = −6 + square root 20, x = −6 − square root 20
2(x + 3)2 = 32; x = 1, x = −7
2(x + 3)2 = 32; x = 13, x = 19

Answer 1

@jim_thompson5910 please help with this one

Answer 2

\[\Large 2x^2+12x-14\] is the same as \[\Large 2x^2+12x+(-14)\]

Answer 3

\[\Large {\color{red}{2}}x^2+{\color{green}{12}}x+({\color{blue}{-14}})\]
is in the form
\[\Large {\color{red}{a}}x^2+{\color{green}{b}}x+{\color{blue}{c}}\]

where
\[\Large \color{red}{a = 2}\]
\[\Large \color{green}{b = 12}\]
\[\Large \color{blue}{c = -14}\]

Answer 4

making sense so far?

Answer 5

i think so...

Answer 6

would it be b?

Answer 7

how did you get to b?

Answer 8

im not sure tbh its kinda a guess

Answer 9

ok let's forget about the answer choices for now. Let's just focus on each step by step

Answer 10

do you see how I got
\[\Large \color{red}{a = 2}\]
\[\Large \color{green}{b = 12}\]
\[\Large \color{blue}{c = -14}\]

Answer 11

okay and we plug that in?

Answer 12

plug a = 2 and b = 12 into h = -b/(2a) to get h = ???

Answer 13

-3?

Answer 14

good

Answer 15

now we'll plug x = -3 back into y = 2x^2+12x-14 to get y = ???

Answer 16

-32

Answer 17

so the x coordinate of the vertex is -3, meaning h = -3
the y coordinate is -32, so k = -32

Answer 18

plug in a = 2, h = -3, k = -32 into
\[\Large y = a(x-h)^2 + k\]
to get
\[\Large y = 2(x-(-3))^2 + (-32)\]
\[\Large y = 2(x+3)^2-32\]

Answer 19

is it d?

Answer 20

Plug in y = 0 and solve for x
\[\Large y = 2(x+3)^2-32\]
\[\Large 0 = 2(x+3)^2-32\]
\[\Large 2(x+3)^2=32\]
\[\Large (x+3)^2=16\]
\[\Large x+3=4 \ \ \text{ or } \ \ x+3 = -4\]
\[\Large x=1 \ \ \text{ or } \ \ x = -7\]

Answer 21

oh so c?

Answer 22

@jim_thompson5910

Answer 23

correct

Answer 24

thank you!

Answer 25

yw

Answer 26

\[2x^2+12x-14=0,2\left( x^2+6x+\left( \frac{ 6 }{ 2 } \right)^2-\left( \frac{ 6 }{ 2 } \right)^2 \right)-14=0\]
\[2\left( x+3 \right)^2-18-14=0,2(x+3)^2-32=0,(x+3)^2-16=0,(x+3)^2-4^2=0\]
(x+3+4)(x+3-4)=0,
(x+7)(x-1)=0
x=-7,1