Question

Algebra Help

Answer 1

\[\sqrt{x+5}-x=-1\]

Answer 2

I know that you first add x on each side. You get \[\sqrt{x+5}=x-1\]

Answer 3

square now

Answer 4

Then you square to get \[x+5=x^2-2x+1\]

Answer 5

subtract one\[3x+4=x^2\] and add 2x on both sides

Answer 6

Now I don't know.

Answer 7

If I may help?

Answer 8

Anyone is welcome to help lol. Please

Answer 9

Are you familiar with the quadratic formula?

Answer 10

Oh, of course... Alright, I think I know how to do it now. Thanks!

Answer 11

Any Help... Any Time...

If you want me to walk with you through it, I don't mind

Answer 12

Is it alright if I quickly solve and ask if it's correct?

Answer 13

Yes, No problem.

Answer 14

Okay, I got x=4 and x=-1
It can't be -1 cause the equation would be false, so it's x=4.

Answer 15

You mean:
\[x^2-3x+4=0\]
\[(x-4)(x+1)=0\]??

Answer 16

Yea. So x would be 4 or -1

Answer 17

And if I plug back in, only 4 works.

Answer 18

I don't think so!
Let me show it to you if you don't mind.

Answer 19

You made a mistake, it should be x^2-3x-4.

Answer 20

a=1
b=-3
c=4

Direct substitution will lead to two imaginary conjugates:\[\frac{ 3 }{ 2 }+\frac{ \sqrt{7} }{ 2 }i\]
and
\[\frac{ 3 }{ 2 }-\frac{ \sqrt{7} }{ 2 }i\]

Answer 21

c is -4

Answer 22

Sorry
It would be easier now
(x+1)(x-4)=0

but why did you exclude -1?

Answer 23

If you plug in -1 to the original equation, you would get 3=-1 which is not true
So x=-1 cannot be an equation.

Answer 24

How "you would get 3=-"

\[x^2-3x-4=(-1)^2-3(-1)-4=1+3-4=0\]
It is correct and satisfy the original equation!

Answer 25

That isn't the original equation. Look all the way at the top :P

Answer 26

-1 only satisfies the quadratic equation.

Answer 27

Anyways, thanks for the help :D
I extremely appreciate it.

Answer 28

I thank you that you gave me that chance to help you!

Answer 29

Thank you for the medals!