Simplifying a Radical Expression in Three Ways

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[res]

Love the commentary ride that accompanies the solution being worked out...I did this in about 15...de-rusting my brain with your truly fantabulous videos! Rock on...

[TedHopp]

I solved it using two other methods.

First method: Multiply numerator and denominator by ✓(x+✓x) - ✓(x-✓x). The numerator collapses to 2✓x and the denominator to ✓(2x).

Second method: Define a = ✓(x+✓x) and b = ✓(x-✓x). Then note that x = (a^2+b^2)/2. Also note that ✓(x^2-x) = ✓((x+✓x)(x-✓x)) = ab. The numerator becomes a+b and the denominator simplifies to ✓((a+b)^2/2), so the whole thing again simplifies to ✓2.

[noone-keol]

Answer is √2 which is approximately equal to 1.414 nice problem.I used the 2nd method

[OdedSpectralDrori]

You can differentiate the expression to show that the derivative is zero before going into method 3, this way you have analytic proof it's going to be a constant

[angelmendez-rivera]

To start with, x >= 0, and furthermore, x – sqrt(x) >= 0, which implies x – sqrt(x) = sqrt(x)^2 – sqrt(x) = sqrt(x)·[sqrt(x) – 1] >= 0, which implies sqrt(x) – 1 >= 0, which implies sqrt(x) >= 1, which implies x >= 1. Also, x^2 – x >= 0, which implies x·(x – 1) >= 0, which implies x >= 1 anyway. Overall, x >= 1. With this caveat in mind, notice that x + sqrt(x) and x – sqrt(x) are conjugate expressions, with [x + sqrt(x)]·[x – sqrt(x)] = x^2 – x. Therefore, sqrt[x + sqrt(x)]·sqrt[x – sqrt(x)] = sqrt(x^2 – x). Also, [x + sqrt(x)] + [x – sqrt(x)] = 2·x. Therefore, sqrt[x + sqrt(x^2 – x)] = sqrt{(sqrt[x + sqrt(x)]^2 + sqrt[x – sqrt(x)]^2)/2 + sqrt[x + sqrt(x)]·sqrt[x – sqrt(x)]} = sqrt[(sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])^2/2] = sqrt[(sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])^2]/sqrt(2) = (sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])/sqrt(2). Therefore, (sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])/sqrt[x + sqrt(x^2 – x)] = (sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])/[(sqrt[x + sqrt(x)] + sqrt[x – sqrt(x)])/sqrt(2)] = sqrt(2).

[sberacatalin]

Super, super misto!!! Excellent exercise! Next?!

[lightfantastic]

Found a neat example based on and inspired by one of your other videos…the second expression is contained in the first which is
2 rt(2+rt(16+8 rt(3)))
Second is
Rt(16+8 rt(3))
Both simplify to
2 + 2 rt(3)
!

[robertlunderwood]

Beware that there are constraints to the value of x. Once you take the denominator and all the radicals into account, x>=1.

[AllanPoeLover]

I think this question have 2 answers. When X not equat to 0, this equation equals to Sqrt(2).
When x = 0, this equation comes to be 0/0, and I think 0/0 equals to 1.

[siddharthabhattacharya]

the first method is brilliant i liked it i went on to rationalise 3-4 times to get the answer

[noahvale]

As soon as I saw this problem I did method in my head. Guess I'm learning from you. Thanks

[vcvartak]

There's is absolutely no difference between 1st and 2nd method. Third method you have proove for x=1, what if x gt 1.

[imranakhtar]

hii sir, how r u, root 2 is the answer, both the methods are explained in a great manner by you, nice !

[maryamhosseini]

What app/software do you use for writing notes?

[leif]

I just squared the whole thing..and got positive and negative square root of 2.

[jarikosonen]

Problem seems to find right substitution... (s or A this case)