A Nice Radical Equation

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Опубликовано:

4 окт 2024

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Комментарии : 27

@guyvankerckhoven 3 дня назад

As a 73 year old retired engineer, I really enjoy these math problems. Thanks SyberMath!

@SyberMath 3 дня назад

Wow! Glad to hear that

@kenclough12345 3 дня назад

minus 2 works by carefully selecting the negative roots

@bobbyheffley4955 3 дня назад

It is a great idea to check radical or rational equations to reveal extraneous solutions.

@sootierr 3 дня назад

General sol. for sqrt(a-sqrt(a-x))=x is x=(-1+sqrt(4a+1))/2

@SyberMath 3 дня назад

That’s a nice problem

@NilestienRamaeuler 3 дня назад

wait isnt golden ratio considered as a solution cuz its value is like 1.618 which compeletly satisfies the domain....the graphs should be intersected at 2 points

@rodrigomarinho1807 2 дня назад

The domain of the function on the left is -2 ≤ x ≤ 2, and it's an injective one. Its maximum is √2.

@scottleung9587 3 дня назад

I used the first method, except that I used long division.

@hassanbagheri8265 3 дня назад

Great. I really enjoyed.

@SyberMath День назад

Glad you enjoyed it

@dan-florinchereches4892 3 дня назад

I didn't know infinitely nested radicals before RU-vid competition problems but they are super useful. Thinking this as nested radical and squaring you get 2-x=x^2 x^2+x-2=0 (X-1)(X+2)=0 nice problems from sybermath

@SyberMath 3 дня назад

Thanks 😍

@othman31415 3 дня назад

We can also solve it by the trig substitution x=2cos(a)

@giuseppemalaguti435 3 дня назад

Per simmetria √(2-x)=x...x=1

@AntonioOlivares-on3fl 3 дня назад

x = 1; sqrt(2-sqrt(2-1))=1.

@pjb.1775 3 дня назад

wooowwwww

@RashmiRay-c1y 20 часов назад

Let 2=a. Then, √(a-x) = a-x^2 > a^2-(2x^2+2)a +(x^4+x)=0 > a = 2 = 1/2[2x^2+1 +/-√(4x^2-4x+1)] = 1/2[2x^2+1 +/-(2x-1)]. Thus, we have x^2+x-2=0 or x^2-x-1=0. But x is positive. So, x=1, 1/2(√5+1). The latter is not a valid solution. So, x=1.

@rakenzarnsworld2 3 дня назад

x = 1

@Nomalum-w7v 2 дня назад

Ajoyib 🇺🇿🤝🇹🇷

@SyberMath День назад

😍

@prollysine 3 дня назад

we get , (x-1)(x^3+x^2-3x-2)=0 , x=1 , x^3+x^2-3x-2=0 , (x+2)(x^2-x-1)=0 , x= -2 , x^2-x-1=0 , x= (1+V5)/2 , (1-V5)/2 , test , x=1 , 1=1 , OK , x=-2 , V2 not -2 false , x= (1+V5)/2 , 1.1755 not 1.61803 false , x= (1-V5)/2 , 0.61803 not -0.61803 false , solu , x=1 ,

@Don-Ensley 3 дня назад

x ∈ { (1-√5)/2, 1, (1+√5)/2} Let u² = 2-√(2-x) Then x = 2-(2-u²)² u = 2-(2-u²)² u⁴-4u² +u+2 = 0 Σ coefficients = 0 u = 1 is is a root factor (u-1) (u³ +u²-3u-2) = 0 u³ +u²-3u-2 = 0 u = -2 is a root by RRT Factor (u+2)(u²-u-1) = 0 u = -2 yields x = -2 which is extraneous (not a working solution) u²-u-1 = 0 u = (1± √5)/2 x = (1± √5)/2 both of these verified to work.

@rob876 3 дня назад

0 x = 0 (x) or x = 1 x = 1

@nasrullahhusnan2289 3 дня назад

sqrt[2-sqrt(2-x)]=x sqrt[{(2-x)+x}-sqrt(2-x)]=x (2-x)+x-sqrt(2-x)=x² x²-x=(2-x)-sqrt(x-2) x²=2-x --> x²+x-2=0 (x+2)(x-1)=0 --> x={-2 1} x=-2 is rejected --> x=1