How do you work square root problems when there is a number outside the radical that needs to be divided out.
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?
16 answers
4 (5x-1) = 9 (2x+4) you mean?
By 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.
so you can introduce answers that do not work in the original.
The 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.
look
5 sqrt (25) = sqrt (625)
5^2 (25) = 625
25 * 25 = 625
625 = 625
5 sqrt (25) = sqrt (625)
5^2 (25) = 625
25 * 25 = 625
625 = 625
How 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
this is all because
[5 sqrt(25) ][5 sqrt(25)] = 25 * 25
[5 sqrt(25) ][5 sqrt(25)] = 25 * 25
WHERE ARE YOU GETTING THE 5 AND THE 25. HOW ARE YOU MULTIPLYING THESE RANDOM NUMBERS TOGETHER?
I 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)
the point is that in your problem
[ 2 sqrt(5x-1) ] [ 2 sqrt (5x-1) ]
= 2*2 * (5x-1)
= 4 * (5x-1)
Perhaps say
(a * b)^2 = a^2 * b^2
whatever a and b are
(a * b)^2 = a^2 * b^2
whatever a and b are
my 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.
You could have told me you made up a completely different problem.
Ok, 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 :)
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 :)
So you distribute the 3 and the 2 to the numbers inside the radical?
Yes, you can do that but not necessary
hen 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.
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.
Thanks 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.
You are very welcome - good luck!
1 answer