I know you have to rewrite the expression into two radical expressions, but there is no square root of either 8 or 2.

Thanks a LOT!

- Thread starter Ladybug
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I know you have to rewrite the expression into two radical expressions, but there is no square root of either 8 or 2.

Thanks a LOT!

It might help to start by doing the multiplication under the radical sign:Ladybug said:

I know you have to rewrite the expression into two radical expressions, but there is no square root of either 8 or 2.

Thanks a LOT!

sqrt(8*2) = sqrt(16)

Now....can you finish?

An alternative method might be to write both 8 and 2 as the product of prime factors:

sqrt(8*2) = sqrt(2*2*2*2)

or, sqrt(2<SUP>4</SUP>)

And then finish.....

Oh dear! Now I feel I was rather...errr...stupid. But they never said I could do it that way in my Algebra book. I didn't realize I could raise 2 to the power of four, either.It might help to start by doing the multiplication under the radical sign:

sqrt(8*2) = sqrt(16)

Now....can you finish?

An alternative method might be to write both 8 and 2 as the product of prime factors:

sqrt(8*2) = sqrt(2*2*2*2)

or, sqrt(24)

And then finish.....

Thanks so much! That made it so clear.

Okay, I won't feel TOO dumb. Thanks for the reassurance! The process of finding the square root of 20 sounds clear. Put the coefficent 2 and then the square root mark with 5 in it. I just learned that not too long ago. So I was baffled when 8 and 4 both didn't have square roots...still, silly that I didn't multiply...Denis said:It's justa learning process, LadyB; DO NOT feel stupid, hear :?:

You can also do stuff like:

sqrt(20) = sqrt(4 * 5) = 2 * sqrt(5) : yepper ??

Thanks for everyone's help! I really appreciate it.

--------------------------------melanie said:solve for x

sqrt(x+3)=X-3

Solve for x :

\(\displaystyle \sqrt{x+3}=x-3\)

\(\displaystyle Square \ both \ sides \ ,\ you'll \ get \ : \\)

\(\displaystyle x + 3 = (x-3)^2\)

\(\displaystyle x + 3 = x^2 - 2(x)(3) + 9\)

\(\displaystyle x^2 - 6x + 9 - x - 3 = 0\)

\(\displaystyle x^2 -7x + 6 = 0\)

\(\displaystyle (x-1)(x-6) = 0\)

. . . . . . . . . . . . .. . I did this for you , but you must show an attempt for any future problem.