Solve Radical (Square Root) Equations with Two Radicals
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- เผยแพร่เมื่อ 9 ก.พ. 2025
- This video shows how to solve square root (radical) equations with two radical expressions. The equations are rearranged so that one radical is on each side of the equals and then both sides are squared. This process is then repeated until there are no radicals remaining.
TYSMMM. I’m in Algebra 3 Honors and trying my best to maintain my 96% in the class. Our teacher never went over a problem like this where one side of the equation has a constant after the square root (3) on problem a. This makes so much senseeee
I'm glad it helped! This double radical problems are some of the most challenging in all of algebra, so you're in good shape if you can get these.
Thank you for this video. Straightforward and easy to understand. Following along helped me get a problem I had been stuck on for nearly 20 minutes, massive gratitude.
Glad I was able to help!
Square both sides. Then simplify. The simplification is very long and should be accurate. It must also follow the laws of exponents.
On (b), it's best to add one of the square roots to the other side before squaring.
@@johnssolutionset yes the radical should be transpose to the other side. It would be more complicated if you dont transpose it.
Nice but quick question
When you have 6√x-2 on LHS and 3x-6 on RHS, from collecting like terms.
Could you divide through by 3 and therefore get 2√x-2= x-2. I did this and then squared both sides. The values were easier to work from then but not sure if the steps i choose is correct?
Dude the fact that i still dont get it makes me want to jump
I'm so done why is math hard
These radical equations are one of the hardest topics, so if you can get through this bit, it eases up a bit. Try this video: th-cam.com/video/nDLPtIMz6gE/w-d-xo.html
So by squaring the radical X-2 +3 will not undo the x-2 under the root ? The very first problem
that's correct - squaring a square root undoes the square root but leaves the x-2 as it is :)
Im gonna cry i still dont get it, how do u do the 2AB part
The 2AB part is one of the most challenging aspects of this whole thing! Here are two videos just focusing on that:
th-cam.com/video/whA4Baz5y-M/w-d-xo.html
th-cam.com/video/OkHMg0H_6xo/w-d-xo.html
Once you master the 2AB part, return to the radical exercises. It should be much easier. Hang in there! Once you get it, things flow.
it is easy
what u have to do is when u expand the (A+B)^2 First A^2 + and then multiply the values inside the brackets by 2 and the again B^2 as the square of the last value
just memorize it then it will be easy.
😀 god bless....
On the equation rule when do you know to use + or subtracting in the beginning for (a+b)^2 =a^2-2ab+b^2
It depends on the sign between the two terms. The two rules are (a+b)^2 =a^2+2ab+b^2 and (a-b)^2 =a^2-2ab+b^2, so note that the sign between the a and b is the same as the sign before the 2ab.
I tried doing this with the equation : sqrt(9-x) = -2 + sqrt(4x -4) and it didnt work... my test is tomorrow 😬
Try rewriting the RHS as sqrt(4x-4) - 2, it should work then. Everything gets mangled when you have a negative in the first term of a binomial. It still works, but you need to be very careful with the signs. It's much easier to just rearrange first.
i love you man
Glad I could help!
Saw thumbnail and guessed squaring immediately
Squaring indeed, but one must be sure to include the "2AB" term in the middle - the biggest mistake with these is not including the 2AB term from (A+B)^2.
I have an one doubt xsquare +24x+144=48x+64 then xsquare-24x+80=0 how? What about 48x after the equal symbol
I subtracted (48x+64) from both sides to get -24x+80 on the left and 0 on the right ☺️ ☺️
Thanks bro I understood
I am hesitat to ask this i have difficulties in division can you please teach division for me.i don't how to divide fast 3 digit and 2 digit can you please teach that please bro .this is my humble request to you
I'm not too good at fast division myself! I'd do some searches on TH-cam and see what comes up. Let me know if you come up with anything useful - I'd love to watch it.
Calculus without rules is not mathematc.
1fest equation is equivalent to :
6sqrt(x-2)=3x-6 equivalent to
4(x-2)=(x-2)^2 and x>=0.
2nd equivalent to :
sqrt(2x-4)+2= sqrt(3x+4).
In this way, we can square both sides because they are positives (same sign).
I agree that it's easier to square when the sign is positive - good point
@user-ky5dy5hl4d True! I was a little sloppy there. It should have indeed said (-6) squared. Thanks for pointing that out!
a) x=2
b) x=4
Conditions for existence of square????? Conditions for this compatibility??????? 2 and 4 it's o.k. in the up conditions???!!! Strange! Very strange!
Halfway there - you're missing solutions in both of these. Both 6 and 2 are answers to (a), and both 4 and 20 are solutions to (b).
@@sberacatalin2250 it's ok - everything's ok here.
I wud've got rid of the factor 3 in the last step :( in (a) .
That would have eventually given the right answer but required just a little more work! ☺
x=4
x=4 is one solution, but you're missing the second solution: x=20
that what i got
Namaste🙏.
🙏😊
a) 6,2....b)12+8i,12-8i..... Ho sbagliato b.... che risulta 4,20
Good catch! I wouldn't be so cruel to include complex answers on this video because it came technically before they are introduced in the class.