Question

I need the solutions of 8= (x+3)^2

Answer 1

Lets first F.O.I.L. the (x+3)^2

Answer 2

We get \[x ^{2} + 6x + 9\]

Answer 3

So we have \[8 = x ^{2}+6x+9\]

Answer 4

Let's subtract the 9 from both sides.

Answer 5

-1 = x^2 + 6x

Answer 6

i need help please help me

Answer 7

Sorry @verity, ask your question elsewhere. I am busy here.

Answer 8

No offense!

Answer 9

@nodive Do you have any questions so far?

Answer 10

nope

Answer 11

Great!!

Answer 12

take the square root of both sides

\[\pm \sqrt{8} = x + 3\]

now subtract 3

\[-3 \pm \sqrt{8} = x\]

the radical can be simplified if necessary....

hope it helps

Answer 13

So from this point. We want to solve for x.

-1 = x^2 + 6x

Let's factor out the x.

-1 = x (x+6)

Answer 14

Sorry. Step back a few steps.

Answer 15

if you distribute and then set the equation to zero, you will need the general quadratic formula to get the solution.

by starting with doing the opposites it avoids unnecessary work.

Answer 16

Omg, I am really sorry, @campbell_st is absolutely right. I apologize so much for walking you through wrong..

Answer 17

g(x)=x^2+6x+1
0=x^2+6x+1
-1=x^2+6x
-1+9=x^2+6x+9
8=x^2+6x+9
8= (x+3)^2

Answer 18

the method @Cosmichaotic has posted is correct... just the long way around

Answer 19

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

Answer 20

This is the quadratic formula. Once you put the equation in ax^2 + bx + c = 0 (Standard Form), you will know where to substitute... I can walk you through this if you like.

Answer 21

8 = (x+3)^2

-> (x+3)^2 = 8
-> x^2 + 6x + 9 = 8
-> x^2 + 6x + 9 - 8 = 0
-> x^2 + 6x + 1 = 0

Where,
a = 1
b = 6
c = 1

Answer 22

Lol, I am really destroying this answer aren't I....

Answer 23

so \[-6\pm 4\]

Answer 24

nope if you use the general quadratic formula its

\[\frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{-6 \pm \sqrt{32}}{2}\]

now simplify

Answer 25

Well done. Sorry about that.

Answer 26

i couldnt have done any better @Cosmichaotic

Answer 27

to simplify it becomes

\[\frac{-6 \pm \sqrt{4 \times 8}}{2} = \frac{-6 \pm \sqrt{4}\times \sqrt{8}}{2} = \frac{-6 \pm 2\sqrt{8}}{2}\]

then you'll get

\[-3 \pm \sqrt{8}\]

as I said... the long way round

Answer 28

thank yall!

Answer 29

One of the solutions presented made no sense. It was this one:

\(8 = (x+3)^{2}\)

Take the Square Root of both sides.

\(\pm \sqrt{8} = x+3\)

This is no good. \(\sqrt{8} > 0 \)

What should have happened was:

Take the Square Root of both sides.

\(\sqrt{8} = \pm (x+3)\)

It may not make much difference as far as the solution is concerned, but it presents a better understanding of the square root operation.

Answer 30

tbh im just lost in general

Answer 31

There are multiple examples. Please read carefully. You should see it.