Question

Simplifying radical expression..

Answer 1

\[\sqrt{a+6}=a\]

Answer 2

I don't see a product property there...
What do you want to do with this equation?

Answer 3

Oh simplifying radical expressions.

Answer 4

The radical expression on the left is as simple as it can get.
Would you like to try solving the equation?

Answer 5

I'm searching everywhere how to do them I just can't get it for some reason.. wasn't there when we learned it please explain the concept of doing it..

Answer 6

I can only assume that given \(\sqrt{a+6}=a\), the only thing worthwhile to do is solve for a.

If that is what you are trying to do, then recall that the inverse of taking a square-root of a number is to square a number.

Does that help?

Answer 7

I hate being behind but thanks just a little. I probably well ask for more help on here I just got mixed up with the assignments I have to do.

Answer 8

Ok, well also make sure that you know what the assignment is actually asking you to do.

You first asked about product property, but there was no product property to be used, then you said simplifying radical expressions, but there was no radical expression to simplify.

I see a radical equation that can be solved, but if you're not even sure that solving the equation is what you're supposed to do then I can't help you.

Answer 9

Yes it's a radical equation except I wrote something totally different it is \[\sqrt{a}+6=32\] radical expresstion

Answer 10

Ok, that's even easier. First isolate the radical by subtracting 6 from both sides.

Answer 11

I got a=676 right? rad(a)=26 and from there I think my techer said to do this \[(\sqrt{a})^{2}=(26)^{2}\]

Answer 12

I really do have problems right now stressing over this to much I did that assignment.. I am on radical equation \[\sqrt{a+6}=a\] i sreally what I have to do and don't understand..

Answer 13

for √a+6=32, you can check your solution of a=676, but substituting that back in:

√676 + 6 = 32. That looks like a true statement to me, so that solution is correct.

For \(\sqrt{a+6}=a\), you can do the same thing. Cancel the square root by squaring both sides.

Answer 14

So what do I do from here?\[a+6=a ^{2}\] -6 -6

Answer 15

Now you have a quadratic equation, \(a^2-a-6=0\)

Answer 16

He said I should usally get two answers I just kinda am guessing a=0 a=6?

Answer 17

What throws me off is the powers.. at the end.

Answer 18

I think I got it thanks for everything

Answer 19

a=0 and a=6 won't work.

This can be factored to (a-3)(a+2)=0.

Answer 20

So a=3 and a=-2 ?

Answer 21

Again, you can check those solutions by putting them back in (one at a time) to the original equation and see if you get a true statement.

e.g. testing a=3 \(\rightarrow \sqrt{(3)+6}=(3)\)
\(\rightarrow \sqrt{9}=3\)

Yeah, that checks out.
Can you check the other one?

Answer 22

\[\sqrt{-2+6}=2\] \[\rightarrow \sqrt{4} = 2\]

Answer 23

Looks good.

Answer 24

Thank you! you are amazing.

Answer 25

And you see that \( (a-3)(a+2)=a^2-a-6 \), right?

Answer 26

Oh yea I forgot about that um not really sure

Answer 27

You'll have to figure that out on your own. Look into how to FOIL two binomials, and how to factor a quadratic trinomial.

Answer 28

So lets say I do n=\[n=\sqrt{2-n}\] then\[n ^{2}=2-n\] and then from there \[n^{2}-n-2=0\] then \[(n+1)(n-2)=0\] Does that look right or no? just seeing if I have it down yet..

Answer 29

or (n-1)(n+2)=0? Idk if that not right then I'm just gonna get some tutoring tomorrow but I appreciate you trying

Answer 30

\(n=\sqrt{2−n}\)
\(\rightarrow n^2=2−n\)
\(\rightarrow n^2+n−2=0\)
\(\rightarrow (n+2)(n−1)=0\)