Radical equation problems?

**homeylova223** · Dec 15th 2010, 02:53 PM

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 (x+2)-7= square root (x+9) Can anyone please help me?

**snowtea** · Dec 15th 2010, 03:17 PM

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

**Archie Meade** · Dec 15th 2010, 03:31 PM

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 (x+2)-7= square root (x+9) Can anyone please help me?

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

$a+12=b$

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

$b=16$

**homeylova223** · Dec 15th 2010, 04:17 PM

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.

**homeylova223** · Dec 15th 2010, 04:18 PM

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

**snowtea** · Dec 15th 2010, 04:20 PM

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 $x$ ?

**homeylova223** · Dec 15th 2010, 05:52 PM

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

**snowtea** · Dec 15th 2010, 05:58 PM

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

**HallsofIvy** · Dec 16th 2010, 02:37 AM

Yes.

**Archie Meade** · Dec 16th 2010, 02:57 AM

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...

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

**homeylova223** · Dec 16th 2010, 09:45 AM

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

**snowtea** · Dec 16th 2010, 09:50 AM

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

**Archie Meade** · Dec 16th 2010, 10:47 AM

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....

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

$A=\sqrt{a+21}$

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

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

$2A-1=9$

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

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