my problem is
square root (a+21)-1= square root(a+12) I am kind of stuck on how to solve this as I do not know what to do with -1.

Also the next equation I need help solving is very similar

2. $\sqrt{a+21}-1=\sqrt{a+12}$
$(\sqrt{a+21}-1)^2=(\sqrt{a+12})^2$
$a+21 - 2\sqrt{a+21} + 1 = a+12$
Use algebra to get $\sqrt{a+21}$ on one side. Once you do that, you can square both sides and no more square roots. Now you can solve for $a$. It will be a quadratic equation.

Note: Remember once you get a solution, plug it back in to check. Especially for square root problems, you can get extra "fake" solutions. Always plug in and check.

3. Originally Posted by homeylova223
my problem is
square root (a+21)-1= square root(a+12) I am kind of stuck on how to solve this as I do not know what to do with -1.

Also the next equation I need help solving is very similar
$\sqrt{a+21}-1=\sqrt{a+12}$

$a+12=b$

$\sqrt{b+9}-1=\sqrt{b}$

$b=16$

4. The back of my book state the answer for the first equation is apparently 4. Also quick question how do you move -2 square root(a+2) to the other side of the equation.

5. I mean -2 square root( a+21) in the work snowtea did.

6. How do you get anything on one side?

Pretend $\sqrt{a+21}$ is just $x$.

I.e. $a+21 - 2x + 1 = a+12$

How do you solve for $x$?

7. would you just add -2 square root(a+21) to the other side of the equation?

8. Yes, can you solve for $x$ in $a+21 - 2x + 1 = a+12$?
You should get $x = ?$, then just replace it with $\sqrt{a+21} = ?$

9. Yes.

10. Originally Posted by homeylova223
would you just add -2 square root(a+21) to the other side of the equation?
No, you add $2\sqrt{a+21}$ to both sides.

You cancel $-2\sqrt{a+21}$ from the side it's on.

If two sides are equal, then if you add the same thing to both sides, they will still be equal.

As in.... 5=5, 5+2=5+2 or 5-2=5-2 etc

therefore

$a+22-2\sqrt{a+21}=a+12$

Add $2\sqrt{a+21}$ to both sides and then subtract $(a+12)$ from both sides.

Alternatively, subtract $(a+22)$ from both sides.
Then divide both sides by $-2$ and square both sides.

You could also go this route...

$\displaystyle\sqrt{a+21}-1=\sqrt{a+12}$

$\displaystyle\sqrt{\left(\sqrt{a+21}-1\right)^2}=\sqrt{a+12}$

$\displaystyle\sqrt{a+21-2\sqrt{a+21}+1}=\sqrt{a+12}$

$\sqrt{a+22-2\sqrt{a+21}}=\sqrt{a+12}$

$a+22-2\sqrt{a+21}=a+12$

$a+12+10-2\sqrt{a+21}=a+12$

which means that

$10-2\sqrt{a+21}=0$

If two values are equal, when we subtract them the answer is zero...

$10=2\sqrt{a+21}\Rightarrow\sqrt{a+21}=5\Rightarrow \ a+21=25$

11. So far this is how far I got

I have
a^2+21-2 square root (a+21)+1= a+12

The I just turn sqrt (a+21) into A
So I get
A^2+21-2(A)+1= a+12

A^2+21+1= 3A+12
Then I would get
A^2-3A+10 I am stuck

12. Take another look at Archie Meads post. It pretty much works out the entire problem.

13. Originally Posted by homeylova223
So far this is how far I got

I have
a^2+21-2 square root (a+21)+1= a+12 there is a typo here on a^2

The I just turn sqrt (a+21) into A
So I get
A^2+21-2(A)+1= a+12 another typo on A^2

A^2+21+1= 3A+12 oops! "A" is not equal to "a"

Then I would get
A^2-3A+10 I am stuck

If $A=\sqrt{a+21}\Rightarrow\ A^2=a+21\Rightarrow\ a+12=A^2-9$

Then....

$\left[\sqrt{a+21}\right]^2-2\sqrt{a+21}+1=a+12$

$A=\sqrt{a+21}$

$A^2-2A+1=A^2-9$

$-(2A-1)=-9$

$2A-1=9$

$A=5\Rightarrow\sqrt{a+21}=\sqrt{25}\Rightarrow\ a+21=25$