Question

solve the equation
\[\sqrt{3w+4}+w=8\]
im not sure what to do with the extra w in this.

Answer 1

\[\sqrt{3w+4}=w+8\]

Answer 2

\[(\sqrt{3w+4)^2}+w=8^2\]

Answer 3

i might be going the wrong way help lol

Answer 4

Subtract the w to the other side.

Answer 5

\[\Large \sqrt{3w+4}+w=8\]


\[\Large \sqrt{3w+4}=8-w\]


\[\Large \left(\sqrt{3w+4}\right)^2=(8-w)^2\]


\[\Large 3w+4=(8-w)^2\]


Do you see where to go from here?

Answer 6

So now it's\[\sqrt{3w+4}=w+8\]

Answer 7

square both sides So: 3w=4=(w+8)^2

Answer 8

Did you keep track of your signs?

Answer 9

im trying to

Answer 10

Now we have 3w+4=\[w ^{2}+16w +64]

Answer 11

Subtract 3w from both sides: 4= w^2+13w+64
Then subtract 4 from both sides: 0= \[w ^{2}+13w+68\]

Answer 12

\[(8+w)^{2}\neq(8-w)^{2}\]

Answer 13

shouldnt be a - should there? the original is all +

Answer 14

They're both postives, so the numbers in the parentheses can be positve or negative. However 13w is positive, so it's positive in both.
Mistake: it's + 60

Answer 15

64-4=60

Answer 16

\[w ^{2}+13w+60=0\]

Answer 17

what you do to one side you do to the other rule got it, i got that answer with you, i was just lost on the extra w and how to input it into everything

Answer 18

I'm sorry i followed the wrong ex. It's (8-w)^2

Answer 19

the original was \[\sqrt{3w+4}+w=8\]

Answer 20

Okay, so \[-w ^{2}-16w+64\]

Answer 21

no, (8-w)^2 = w^2 - 16w + 64

Answer 22

then subtract 3w and 4 to get \[-w ^{2}-19w+60\]. Carry it over to the other side so it equals \[w ^{2}+19w-60\]

Answer 23

<~~~~is really confused now lol

Answer 24

Me too..

Answer 25

It's probably best to start over then

Answer 26

lol

Answer 27

\[\sqrt{3w+4}+w=8\]
original equation

Answer 28

\[\sqrt{3w+4}=8-w\]\[3w+4=(8-w)^{2}\]\[3w+4=64-16w+w ^{2}\]\[w ^{2}-19w+60=0\]

Answer 29

Use quadratic formula to solve.

Answer 30

Or factor.

Answer 31

(w-15)(w-4)=0

Answer 32

w=15
w=4
Check in original to verify.

Answer 33

ok im following i had to re-write so i wasnt getting lost

Answer 34

Good, factoring is always easier than using the quadratic, provided it is factorable lol.

Answer 35

true

Answer 36

omg ty, i so got lost there for a few