Question

_____
√x+10=x-2 x+10 is all together under √ symbol

Answer 1

\[\sqrt{x+10}=x-2\]
(Take note: For \(\sqrt{x+10}\) to be defined, you must have \(x\ge-10\).)
Start off by squaring both sides:
\[\left(\sqrt{x+10}\right)^2=(x-2)^2\\
x+10=(x-2)^2\\
x+10=x^2-4x+4\\
x^2-5x-6=0\]

Answer 2

is the next line
0=-6x+1x

Answer 3

then zero principle property?

Answer 4

You're right about the zero product property, but not with what you wrote. I think you meant to factor:
\[(x-6)(x+1)=0\]

Answer 5

hey will you give me a problem to see if i get it on each step....only if im wrong you correct me

Answer 6

but on that one the ultimate answer is x=1 and x=6?

Answer 7

Close: x=-1 and x=6

Answer 8

yes i need to pay closer attention to pos and neg

Answer 9

!hey will you give me a problem to see if i get it on each step....only if im wrong you correct me

Answer 10

Try this one
\[\sqrt{x-1}=-(x+1)\]

Answer 11

x-1=(x+1)^2

Answer 12

Yes, so far so good

Answer 13

x-1=x^2+2x-1

Answer 14

+1 at the end of the right side

Answer 15

x+1=0 and x+1=0?

Answer 16

\[x-1=x^2+2x+1\\
x^2+x+2=0\]
Actually, it turns out you can't factor this easily. Have you learned the quadratic equation?

Answer 17

yes

Answer 18

hold one moment....let me try work this one out on paper

Answer 19

dude you have a negative one on the outside of the right side of radical eq?

Answer 20

im lost on this one

Answer 21

can we try a diffrent one

Answer 22

Ah, sorry I got ahead of myself. There is no (real) solution to the equation I gave you.

Try this one instead:
\[x-3=\sqrt{5x+11}\]

Answer 23

ok 1 sec

Answer 24

0=x^2-11x-20

Answer 25

Wait, something's not right... I must have made a mistake while coming up with the equation...

Answer 26

but my steps were right though werent they

Answer 27

Yes they were. I'll figure out what went wrong on my end.

Answer 28

no need

Answer 29

instead come up with new one

Answer 30

maybe sq rt4x+1is what u meant

Answer 31

No, I had something else in mind. It was supposed to be \(x+3\) on the left side.

Answer 32

Anyway, here's another:
\[\sqrt{x+4}=x+2\]

Answer 33

x+3= square rt of 4x+11

Answer 34

ok

Answer 35

x=0 and x=-3?

Answer 36

Right, but check those solutions.

Answer 37

Only one of them works.

Answer 38

-3

Answer 39

\(x=-3\) isn't a real solution. Can you see why?
\[\sqrt{-3+4}=-3+2~?\]

Answer 40

hold on

Answer 41

yes i see why

Answer 42

lets try another

Answer 43

\[x+6=\sqrt{-x}\]
this one also involves a solution that doesn't quite work.

Answer 44

one moment

Answer 45

x^2+13x+36=0

Answer 46

You can factor that

Answer 47

one moment

Answer 48

x=-9 only checks out to be true the other was -4 but is false

Answer 49

Are you sure about that?
\[-9+6=\sqrt{-(-9)}\\
-3=\sqrt9\\
-3\not=3\]
\(x=-9\) is not a real solution.

\[-4+6=\sqrt{-(-4)}\\
2=\sqrt4\\
2=2\]
\(x=-4\) is a real solution.

Answer 50

what about this problem