In the first step sqrt(3x-5)=2-sqrt(x-1): Left side is positve bcse sqrt() >=0 So the inequality does make sense ONLY if right side has same sign Therefore, 2-sqrt(x-1)>=0 ===> x
Sqrt(A)+sqrt(B)= p , p>0 Conditions : A>=0 ( or B>=0) p^2 >= A+B 4AB= [ p^2-A-B ]^2 Ex: sqrt(x-1)+sqrt(2x-1) = 5 x-1>=0 ==> x>=1 25 >= 3x-2 ==> x x=5 or x=145 145 ejected as not in D.
The square root is the non-negative value, that multiplied with itself is the radicant. If the square root had two results, it would not have a unique result so it would not be a math function. Anyway, the equation x^2 = 4 has both + and - square root of 4 as solutions.
10 works too if you recall that the square root of 9 can be -3. Root 25 + Root 9 = 2 If we choose +5 for the first root and -3 for the second root. 5 - 3 = 2
It would be great if you could explain where the extra solution came from. For those that don't know it is because the square root of a positive number has two solutions, one positive, one negative. The original question only uses the positive solution. However, when you square it you lose that distinction. So, X=10 works if the second square root used the negative root, as you then get 5-3 instead of 5+3.
it's late but if I recall correctly the square root (with the radical) of a positive number is the absolute value of the root, so you can't have a negative solution. however, this is a quadratic equation when you can in theory use both solutions, but you can't use the negative root of the number because the function dictates that you should use its positive root, so according to the rules x = 10 is invalid.
I made substitution t=✓(x-1), that means t>0 and x=t²+1. Equation for t easily derived t²+2t-3=0. t=1 is the only positive root of this equation, so x=2.
Of course if the sum of two square roots is 2 a simple first step is to check for 1 + 1 or 2 + 0. The solution x = 2 appears and as x - 1and 3x -5 are increasing functions further solutions are impossible.
In minute 1:25, you added something in the right. Is that a formula? Can I use it with all equations similar to this one? And thank you!! This helped me lot!!
You are very welcome Hirak ji! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
√n≥0 so √3x-5 + √x-1 = 2 there's one possible is 1+1 = 2 mean √3x-5 = 1,√x-1 = 1 ,x=2 so you can solve it in easy ways! The problems always have many ways to solve.😃
You are very welcome Ikee! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
An 1-dimension equation should have just one single solution. The "-3" coming out of sqrt(9) as raised by several readers actually is introduced to the equation by math calculation process. Therefore, "10" may not be a real solution of the equation!
Dear SoU RaV, I'm not sure your question is: 60/5x =60/6x +12/60 (x is in the denominator) OR (60/5)x =(60/6)x +12/60 (x is in the numerator). Any way, I'll show you both scenarios: 1st scenario: 60/5x =60/6x +12/60 => reduce the fractions => 12/x =10/x +1/5 => Least Common Denominator LCD is: 5x => multiply it by 5x => 5x(12/x =10/x +1/5) => 60=50+x => x=10 is our solution! 2nd scenario: (60/5)x =(60/6)x +12/60 => reduce the fractions => 12x=10x +1/5 => 2x=1/5 => x=1/10 is our solution! Thanks for asking. All the best :)
You are very welcome Michelle! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
You are very welcome Chase! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
Thanks for the feedback! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Daunting at first sight, then logic because of all your previous videos.
Thanks very much Mr.
Happy to help you Wim! You are awesome.
Take care dear and stay blessed😃
Sir I love you. You’re my lifesaver
I was stuck here for 4 hours😭 thank you sooo muchhh. You are a lifesaver 😭♥️
In the first step sqrt(3x-5)=2-sqrt(x-1):
Left side is positve bcse sqrt() >=0
So the inequality does make sense ONLY if right side has same sign
Therefore, 2-sqrt(x-1)>=0 ===> x
Sqrt(A)+sqrt(B)= p , p>0
Conditions :
A>=0 ( or B>=0)
p^2 >= A+B
4AB= [ p^2-A-B ]^2
Ex: sqrt(x-1)+sqrt(2x-1) = 5
x-1>=0 ==> x>=1
25 >= 3x-2 ==> x x=5 or x=145
145 ejected as not in D.
Good .
Sir please why did you move the positive radical to the right hand side
10 is also a solution. (-3) is also the square root of 9, so 5 + (-3) = 2.
Thanks Jeff for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
The square root is the non-negative value, that multiplied with itself is the radicant. If the square root had two results, it would not have a unique result so it would not be a math function. Anyway, the equation x^2 = 4 has both + and - square root of 4 as solutions.
Correct.
Wow! I have been thinking how to solve this for more than 2hr
OMG THANK YOU T^T it's almost Midnight in my place and i was having a hard time but now i think i finally understand it
Nice I had a problem solving these kind of equations nice job 🙂
Glad it helped!
Thanks RS for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard
This envolves exponentials ans radicals ,
Can this equation be Solved using Logarithm ? If yes ,how ?
For the x = 10 solution, you get
√25 + √9 = 2 ... You forget that the roots of 9 are 3 and -3, so 5 + (-3) = 2 is a valid solution.
But square root cannot be negative.
@@TheoSin not so. Whenever you see the square root (or any even root) you need to put a +/- in front of the answer.
Sqrt(9) = +3 or -3
10 works too if you recall that the square root of 9 can be -3. Root 25 + Root 9 = 2
If we choose +5 for the first root and -3 for the second root. 5 - 3 = 2
Yeah, I was about to say...
Very helpful video 👍🏼
It would be great if you could explain where the extra solution came from.
For those that don't know it is because the square root of a positive number has two solutions, one positive, one negative. The original question only uses the positive solution. However, when you square it you lose that distinction.
So, X=10 works if the second square root used the negative root, as you then get 5-3 instead of 5+3.
it's late but if I recall correctly the square root (with the radical) of a positive number is the absolute value of the root, so you can't have a negative solution. however, this is a quadratic equation when you can in theory use both solutions, but you can't use the negative root of the number because the function dictates that you should use its positive root, so according to the rules x = 10 is invalid.
I made substitution t=✓(x-1), that means t>0 and x=t²+1. Equation for t easily derived t²+2t-3=0. t=1 is the only positive root of this equation, so x=2.
Answer x =2
(3x-5)^1/2 + (x-1)^1/2 =2
3x-5 + 2(3x^2 - 8x + 5)^1/2 + x-1 =4 (square both sides)
4x-6 + 2 (3x^2- 8x +5)^1/2 =4
2(3x^2-8x+5)^1/2= 10 - 4x (substract 4x-6 from both sides)
(3x^2- 8x+5)1/2 = 5 - 2x (divided both sides by 2)
3x^2 - 8x + 5 = 25- 20x + 4x^2 (square both sides)
0 = x^2 - 12x + 20 (substract 3x^2- 8x + 5 from both sides)
0 = (x-2)(x-10)
2 = x (from x-2)
10 =x (from x-10)
2 and 10 = x
Thanks my friend for prompt input👍
It's better !
Before square both sides of sqrt(3x^2-8x+5)=5-2x
Include restriction: 5/3
Of course if the sum of two square roots is 2 a simple first step is to check for 1 + 1 or 2 + 0. The solution x = 2 appears and as x - 1and 3x -5 are increasing functions further solutions are impossible.
Thanks Geoffery for the feedback. You are awesome 👍 Take care dear and stay blessed😃
7:44
Actually sir th value of √9 could also be -3
In that case 5-3=2
The value could not be negative under the square root
In minute 1:25, you added something in the right. Is that a formula? Can I use it with all equations similar to this one? And thank you!! This helped me lot!!
Lhlk
Kgignto455u7920ogkhhavgkbmf hdfhglkelhlelofiyuyiTkwl3OEKYlelhoosjuh6ihuh6h6jj5u5j4ki4io2otp2moiysfzvcgr21etiohkmbd bg6euooiypyitpgou
Thank you sir..
You are very welcome Hirak ji! Thank you so much for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep sharing my channel with your family and friends. Take care dear and all the best 😃
@@PreMath of course sir...l 'll keep sharing it...and thank you sir ...ur teaching method is very understanding...
Where did ,-2(2) came from
2. Based on inspection. Easy.
Where is the checking? If your answer is correct
How we Identify the answer is correct?
Thank you for answering my questions! Advance
At the end of the video....
Where were you when I was a teenager 😢🤨🤗
√n≥0 so √3x-5 + √x-1 = 2 there's one possible is 1+1 = 2 mean √3x-5 = 1,√x-1 = 1 ,x=2 so you can solve it in easy ways! The problems always have many ways to solve.😃
Thank you💕💕💕
You are very welcome Ikee! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
An 1-dimension equation should have just one single solution. The "-3" coming out of sqrt(9) as raised by several readers actually is introduced to the equation by math calculation process. Therefore, "10" may not be a real solution of the equation!
Got the exact same question on my online quiz lol
Mr, I think you muat check 10 ia also the solution. Check it, ✓25 + ✓ 9 = ( +/- 5) + ( +/- 3)=? You must take +5 -3 = 2. ( So 10 is the solution)
Did anyone else look at the problem and know immediately the answer is x=2?
Sir please solve this 60/5x =60/6x +12/60
Dear SoU RaV, I'm not sure your question is: 60/5x =60/6x +12/60 (x is in the denominator) OR (60/5)x =(60/6)x +12/60 (x is in the numerator). Any way, I'll show you both scenarios:
1st scenario: 60/5x =60/6x +12/60 => reduce the fractions => 12/x =10/x +1/5 => Least Common Denominator LCD is: 5x => multiply it by 5x => 5x(12/x =10/x +1/5) => 60=50+x => x=10 is our solution!
2nd scenario: (60/5)x =(60/6)x +12/60 => reduce the fractions => 12x=10x +1/5 => 2x=1/5 => x=1/10 is our solution!
Thanks for asking. All the best :)
Thank you sir
A tad more easy if substitute y^2=x-1
That certainly works. It even gives a second solution of x = 10 if you accept positive or negative roots
thank you so much for this!!!
You are very welcome Michelle! I'm sure you are a brilliant and bright student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
hi Mr. Sharif
x variable
omg thank u my teacher made no sense
You are very welcome Chase! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
The solving method is just like that of a normal algebraic equation. Nothing too complicating
Thanks for the feedback! I'm sure you are an awesome student 👍 I'm glad you liked it! Please keep sharing my channel with your family and friends. Take care dear and all the best😃
👊😃👍
x=2.
x = 2
Calculus without rules !!!
OK?
0