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January 27, 2015

January 27, 2015

Posted by **Catey** on Monday, March 31, 2008 at 7:41pm.

Example: 2(square root of 5x-1)=3(square root of 2x+4). How can you get both of the radicals by themselves in order to solve the equation?

- Algebra -
**Damon**, Monday, March 31, 2008 at 7:44pm4 (5x-1) = 9 (2x+4) you mean?

- Algebra -
**Damon**, Monday, March 31, 2008 at 7:46pmBy the way be careful to check results whenever you square both sides like that. remember -1^2 = +1^2

so you can introduce answers that do not work in the original.

- Algebra -
**Casey**, Monday, March 31, 2008 at 7:51pmThe two and the three are both being multiplied to the square roots that are on both sides of the equal sign. I don't know how to isolate the radicals(to get everything being added, subtracted, multiplied, or divided away) in order to square everything to get rid of the radicals. If the problem was 4+(square root of 4+3x)=6, I know that all you do is subtract the four at the beginning and you square everything. It would be 4+3x=4 and then solve from there, which this problem probably doesn't work b/c I made it up.

- Algebra -
**Damon**, Monday, March 31, 2008 at 7:54pmlook

5 sqrt (25) = sqrt (625)

5^2 (25) = 625

25 * 25 = 625

625 = 625

- Algebra -
**Casey**, Monday, March 31, 2008 at 7:57pmHow are you getting the five? I am not really understanding. Since the numbers 2 and 3 are being MULTIPLIED, how can you just add them instead of dividing

- Algebra -

- Algebra -
- Algebra -
**Damon**, Monday, March 31, 2008 at 7:58pmthis is all because

[5 sqrt(25) ][5 sqrt(25)] = 25 * 25

- Algebra -
**Casey**, Monday, March 31, 2008 at 8:00pmWHERE ARE YOU GETTING THE 5 AND THE 25. HOW ARE YOU MULTIPLYING THESE RANDOM NUMBERS TOGETHER?

- Algebra -
**Damon**, Monday, March 31, 2008 at 8:00pmI just made the problem with 5 up to show what is happening

the point is that in your problem

[ 2 sqrt(5x-1) ] [ 2 sqrt (5x-1) ]

= 2*2 * (5x-1)

= 4 * (5x-1)

- Algebra -
**Damon**, Monday, March 31, 2008 at 8:03pmPerhaps say

(a * b)^2 = a^2 * b^2

whatever a and b are

- Algebra -
**Casey**, Monday, March 31, 2008 at 8:06pmmy teacher told me that the only way you can square EVERYTHING was if you ISOLATE the radicals. What you're doing is something completely different.

You could have told me you made up a completely different problem.

- Algebra -
**Damon**, Monday, March 31, 2008 at 8:11pmOk, isolate the radicals

2 sqrt (5x-1) is

sqrt [(4)(5x-1) ]

= sqrt ( 20 x -4)

and on the other side

3 sqrt (2x+4) is

sqrt[(9)(2x+4)]

= sqrt (18 x + 36)

then square both sides. It is harder but comes out the same :)

- Algebra -
**Casey**, Monday, March 31, 2008 at 8:16pmSo you distribute the 3 and the 2 to the numbers inside the radical?

- Algebra -
- Algebra -
**Damon**, Monday, March 31, 2008 at 8:17pmYes, you can do that but not necessary

- Algebra -
**Damon**, Monday, March 31, 2008 at 8:21pmhen your teacher said "isolate the radicals, I think the teacher meant "isolate the term with the radical"

like something + sqrt someting is bad

but something times sqrt something is ok

By the way, since further down you said you know how to foil, you know how to square (a + sqrt b)

(a+sqrt b)(a+sqrt b) = a^2 + 2 a sqrt b + b

however your teacher said do not do that because it did not help, you still have a square root in there.

- Algebra -
**Casey**, Monday, March 31, 2008 at 8:27pmThanks for your help, you spent a lot of time with my problem. It is GREATLY APPRECIATED and it helps with the confusion I had on the quiz today. My teacher was not there to answer my question and she had never taught us how to solve these type of problems. Thanks again.

- Algebra -
- Algebra -
**Damon**, Monday, March 31, 2008 at 8:32pmYou are very welcome - good luck!

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