Question

Please help me MEDAL REWARDED!
find the solution to the quadratic equation by completing the square.

@ilfy214 @nincompoop @jdoe0001 @Hero @ganeshie8 @satellite73

Answer 1

\[x^{2}+6x=7\]

\[x^{2}+12x=-11\]

Answer 2

2 equations

Answer 3

yea i know

Answer 4

ok

Answer 5

Lets start with the first equation.

x² + 6x = 7

The first step would be to find out what value you add to both sides. You take your "x" term divide it by 2 and square it.

(6/2)^2
3^2
9

We would add this to both sides of the equation

x² + 6x + 9 = 7 + 9
x² + 6x + 9 = 16

Can you take it from there? Or do I need to continue?

Answer 6

@Dragondd320 you lost me lol take it step by step please

Answer 7

the second one is similar to the first

Answer 8

\[x^2+12x=-11\] think "half of 12 is 6, and six squared is 36
go right to
\[(x+6)^2=-11+36=25\]

Answer 9

when i did it i got -7 and 1 and idk if im corrdct

Answer 10

correct*

Answer 11

Okay, first state your equation.

x² + 6x = 7

Now to solve by completeing the square you use a special formula: (b/2)^2

Answer 12

then you get
\[x+6=5\] or \[x+6=-5\] so
\[x=-1\] or \[x=-11\]

Answer 13

b = 6

so
(6/2)^2
3^2
9

Answer 14

ok @Dragondd320 i understand so far

Answer 15

Now you add this to both sides.

x² + 6x +9 = 7 + 6
x² + 6x + 9 = 13

The left side is a perfect square polynomial. You can factor that down like so:

x² + 6x + 9 = 13
(x + 3)(x + 3) = 13
(x + 3)² = 13

Now I ask are you wanting this solved for x, or just in vertex form?

Answer 16

some typo here unfortunately

Answer 17

what?

Answer 18

add 9 get
\[x^2+6x+9=7+9=16\] or
\[(x+3)^2=16\] other not \((x+3)^2=13\) it was just a typo is all

Answer 19

I'll assume you are looking for the solution to x.

*You are right @satellite73*

(x + 3)² = 16

Answer 20

ok can we start over because im confuse

Answer 21

But I continue.

(x + 3)² = 16

Take the square root of both sides.

(x + 3)² = 16
x + 3 = ±4

Then solve:

x + 3 = ± 4
x = 3 ± 4

So x = 3 ± 4
or:

x = 3+4
x = 7

x = 3-4
x = -1

x = 7 or -1

Answer 22

what goes before what you just wrote because i am confused @Dragondd320

Answer 23

lets solve this one quickly, then maybe it will be clear how it works
\[x^2+8x=9\] half of 8 is 4 and 4 squared is 16, go right to
\[(x+4)^2=9+16=25\] or just
\[(x+4)^2=25\]

Answer 24

taking the square root of both sides, you get
\[x+4=5\iff x=1\] or
\[x+4=-5\iff x=-9\]

Answer 25

im confused on how you started that

Answer 26

@Dragondd320

Answer 27

Sorry, I got sidetracked with something.

What part has you confused?

Answer 28

the first and second equation that you ans @satellite73 did im confused becuase theres was different things being said can you show me how there were suppose to be written from beginning please @Dragondd320

Answer 29

Step 1: State your equation

x² + 6x = 7

Step 2: Complete the square

(b/2)²
(6/2)²
3² = 9

Add this to both sides

x² + 6x + 9 = 7 + 9
x² + 6x + 9 = 16

Step 3: Factor your perfect square

x² + 6x + 9 = 16
(x + 3)(x + 3) = 16
(x + 3)² = 16

Step 4: Square both sides

(x + 3)² = 16
x + 3 = ±4 (the is ± because it could be positive or negative)

Step 5(final): Solve for x

x + 3 = ±4
x = 3 ±4

so

x = 3 + 4 and x = 3 - 4

x = 3 + 4 = 7
x = 3 - 4 = -1

x = 7 and -1

How is that for clearing up purposes?

Answer 30

yes and what about the second one?

Answer 31

Okay follow the same steps.

Step 1: State your equation

x² + 12x = -11

Step 2: Complete the square

(b/2)²
(12/2)²
6² = 36

Add this to both sides

x² + 12x + 36 = -11 + 36
x² + 12x + 36 = 25

Step 3: Factor your perfect square

x² + 12x + 36 = 25
(x + 6)(x + 6) = 25
(x + 6)² = 25

Step 4: Take the square-root of both sides.

(x + 6)² = 25
x + 6 = ±5 (Could be positive or negative remember that when doing square roots)

Step 5: Solve for x

x + 6 = ±5
x = 6 ± 5

so

x = 6+5 and x = 6-5

x = 6+5 = 11
x = 6-5 = 1

x = 11 or 1

Does that all make sense?

Answer 32

yes thank you sooo much

Answer 33

No problem ^.^ glad to have helped