How to Solve Radical Equations that have Two Radicals - Simple Method

[wimm]

Daunting at first sight, then logic because of all your previous videos.
Thanks very much Mr.

[jeffgray]

10 is also a solution. (-3) is also the square root of 9, so 5 + (-3) = 2.

[touhami]

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<=5.

That's why x=10 cannot be solution because 10 >5.

Thanks

[ThomasShelby]

I was stuck here for 4 hours😭 thank you sooo muchhh. You are a lifesaver 😭♥️

[touhami]

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<= 9
Therfore, D: [ 1, 9 ]
4(x-1)(2x-1)= (27-3x)^2 ==> x=5 or x=145
145 ejected as not in D.

[agentmischief]

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

[basirubuwaro]

Wow! I have been thinking how to solve this for more than 2hr

[rssl]

Nice I had a problem solving these kind of equations nice job 🙂

[Paul-sjdb]

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.

[PotPoet]

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

[gloriaoppong]

Sir please why did you move the positive radical to the right hand side

[stvcia]

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.

[marcellanthier]

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.

[geoffreyparfitt]

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.

[devondevon]

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

[nitingl]

This envolves exponentials ans radicals,
Can this equation be Solved using Logarithm ? If yes, how ?

[alexajudithperez]

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!!

[johnyy]

√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.😃

[walterwen]

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!

[glaciveestudios]

The solving method is just like that of a normal algebraic equation. Nothing too complicating