A Nice Radical Math Problem | Can You Solve This?

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A Nice Radical Math Problem | Can You Solve This?
Welcome to infyGyan!
In this video, we explore an intriguing algebra problem involving radicals, perfect for those preparing for Math Olympiad. This question will enhance the understanding of radical expressions and skills of problem-solving. Watch as we break down the solution step-by-step, providing clear explanations and insights along the way.
If you're a Math Olympiad participant or simply enjoy tackling competitive math problems, this video is for you. Make sure to like, subscribe, and hit the notification bell to stay updated with more exciting math challenges. Let's solve this radical algebra problem together!
🔍 In this video:
Detailed walkthrough of a challenging algebra problem.
Tips and tricks for solving complex radical expression.
Encouragement to enhance your problem-solving skills and mathematical thinking.
📣 Call to Action:
Have a go at the problem yourself before watching the solution!
Share your solutions and approaches in the comments below.
If you enjoyed this challenge, give it a thumbs up and subscribe for more intriguing math problems!
🔗 Useful Links:
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Опубликовано:

15 сен 2024

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Комментарии : 19

@michaeldoerr5810 10 дней назад

That is one if the easiest problems under infyGyan's instruction. I will use that for practice!!!

@ManojkantSamal 10 дней назад

Respected Sir, Pranam...... The way you took the problem is much more convinced......

@Shobhamaths 10 дней назад

x^6=9-20/√5; 1/x^6=9+20/√5; x^6+1/x^6=18; (x^2) ^3+(1/x^2) ^3=18; (x^2+1/x^2) *(x^4-1+1/x^4) =18; Let t=x^2+1/x^2 then t^3-3t-18=0 t=3 others are complex x^2+1/x^2=3 x+1/x=±√5🤙

@ManojkantSamal 10 дней назад

^= read as to the power *=read as square root As per question X^6=9-(20/*5)=9-(4.*5) 1/x^6=1/(9-4.*5)=9+4.*5 So, X^6+(1/x^6)=9-4.*5+9+4.*5=18 So, {X^3+(1/x^3)}^2- 2×{x^3×(1/x^3)=18 {X^3/(1/x^3)}^2=18+2=20 X^3+(1/x^3)=*20=2×*5 {(X+(1/x)}^3-(3×

@user-kt1dm9jz5t 10 дней назад

E=(5)^(1/2).

@tejpalsingh366 10 дней назад

?= √5

@abcekkdo3749 10 дней назад

E=√5

@maherom1 6 дней назад

Please résolve this équation sqrt (x-1) +4 sqrt(y-4) = (x+y)/2

@user-kp2rd5qv8g 10 дней назад

9-20/√5 = 9 - 4√5 = (√5-2)^2 > x = (√5-2)^1/3. Now, (√5-1)^3 = 8(√5-2) > √5-2 = [1/2(√5-1)]^3 > x = 1/2(√5-1) > 1/x= 1/2(√5+1) > x+1/x = √5.

@RajeshKumar-wu7ox 10 дней назад

X+ 1/x =√5

@yakupbuyankara5903 10 дней назад

-(5^(1/2)).

@Fjfurufjdfjd 10 дней назад

Ειναι χ=[9-4(5)^(1/2)]^(1/6)=[(5)^(1/2)-2]^(1/3).αρα χ^3= ριζα5-2 και 1/(χ^3)=ριζα 5+2. Αν χ+(1/χ)=y τοτε y^3-3y-2ριζα 5=0. Στο R y= ριζα 5 Ε=5^(1/2)

@ABHISHEKKUMAR-01024 10 дней назад

Let ³√(√5 - 2) = a ...(1) Clearly, a > 0 Also let ³√( - √5 - 2) = b ...(2) Then, a³ = √5 - 2 And, b³ = - √5 - 2 Then, a³ + b³ = - 4 ...(3) Also, ab = { ³√( √5 - 2) } { ³√( - √5 - 2) } = ³√( √5 - 2)( - √5 - 2) = ³√( 2 - √5)( √5 + 2) = ³√( 2² - ( √5)² ) = ³√( 4 - 5) = ³√ - 1 = - 1 ...(4) Now, (a + b)³ = a³ + 3a²b + 3ab² + b³ = (a³ + b³) + (3a²b + 3ab²) = (a³ + b³) + 3ab(a + b) = ( - 4) + 3( - 1) (a + b) = ( - 4) - 3(a + b) ...(5) Let us put a + b = z in (5) : z³ = - 4 - 3z or, z³ + 3z + 4 = 0 ...(6) We observe that ( - 1)³ + 3( - 1) + 4 = ( - 1) + ( - 3) + 4 = - 4 + 4 = 0 Therefore, z = - 1 or, a + b = - 1 ...(7) is a solution. We quit here our investigation of further roots. Also, a + b = - 1 and ab = - 1 implies that a, b are roots of the equation u² + u - 1 = 0 It's roots are u = { - 1 ± √(1² - 4(1)( - 1) } / 2 = { - 1 ± √(1 + 4) } / 2 = (- 1 ± √5) / 2 Here a is one of the values of u and surely a = ( - 1 + √5) / 2 = ( √5 - 1) / 2 Therefore, x = a = ³√( √5 - 2) = ( √5 - 1) / 2 Further x + (1 / x) = {(√5 - 1) / 2} + [ 1 / { (√5 - 1) / 2 } ] = {(√5 - 1)/2} + {2 / (√5 - 1)} = {(√5 - 1)/2} + {2(√5 + 1) / (√5 - 1)(√5 + 1)} = {(√5 - 1)/2} + {2(√5 + 1) / 4} = {(√5 - 1)/2} + {(√5 + 1)/2} = √5 We should suggest the method of rationalization (which is more correct and appropriate) to our kids for the evaluation of x + (1 / x) in these type of cases as we are guiding them for Olympiad and many Talent Search competitions. We have to remain very aware and careful for them!

@RealQinnMalloryu4 10 дней назад

(x^6)^2 ➖ (9)^2 ➖ 20/5= {x^36 ➖ 81} ➖20/5=45 ➖ (20)^2/5 = {45 ➖ 200}/5= 165/5=33 (x ➖ 33x+33 ). {x+x ➖ }{1+1 ➖},/{x+x ➖ }=:{x^2+2}/x^2=2x^2/x^2 =2 (x ➖ 2x+2).

@user-kp2rd5qv8g 10 дней назад

Here, x^6= 9-4√5 > 1/x^6 = 9+4√5 > x^6+1/x^6= 18. Let t=x+1/x. Then, x^3+1/x^3= t(t^2-3) > x^6+1/x^6 +2 = t^2(t^2-3)^2 = a^3-6a^2+9a, where t^2=a. So, a^3-6a^2+9a-20=0 and a=5 is the only real solution. So (x+1/x)^2 = 5. But x+1/x is positive. So, x+1/x=√5.

@vcvartak7111 10 дней назад

It's more direct approach than his manipulations.

@SidneiMV 10 дней назад

x = (9 - 20/√5)^(1/6) x = (9 - 4√5)^(1/6) 9 - 4√5 = (√5 - 2)² => x = ∛(√5 - 2) (√5 - 2)(√5 + 2) = 1 => 1/x = ∛(√5 + 2) x + 1/x = a = ∛(√5 - 2) + ∛(√5 + 2) a³ = (√5 - 2) + (√5 + 2) + 3a a³ = 2√5 + 3a a³ - 3a - 2√5 = 0 a³ - 5√5 - 3a + 3√5 = 0 a³ - √5³ - 3(a - √5) = 0 (a - √5)(a² + a√5 + 2) = 0 a - √5 = 0 a = *x + 1/x = √5*

@walterwen2975 10 дней назад

A Nice Radical Math Problem: x = ⁶√(9 - 20/√5), x + 1/x =? x = 9 - 20/√5 = 9 - 4√5 > 0; x ϵ R; No complex or imaginary value x = 9 - 4√5 = 5 - 4√5 + 4 = (√5)² - 2(2)√5 + 2² = (√5 - 2)² x = ⁶√(9 - 20/√5) = ⁶√(9 - 4√5) = ⁶√[(√5 - 2)²] = ³√(√5 - 2) x + 1/x = ³√(√5 - 2) + 1/[³√(√5 - 2)] = ³√(√5 - 2) + ³√(√5 + 2)/³√[(√5 - 2)(√5 + 2)] = ³√(√5 - 2) + ³√(√5 + 2)/³√(5 - 4) = ³√(√5 - 2) + ³√(√5 + 2) Let: a = ³√(√5 + 2), b = ³√(√5 - 2); a³ + b³ = 2√5, ab = 1, x + 1/x = a + b; a, b ϵ R a³ + b³ = (a + b)(a² - ab + b²) = (a + b)[(a + b)² - 3ab], (a + b)³ - 3(a + b) = 2√5 Let: m = a + b > 0; (a + b)³ - 3(a + b) - 2√5 = m³ - 5√5 - 3m + 3√5 = 0; m ϵ R [m³ - (√5)³] - 3(m - √5) = (m - √5)[(m² + m√5 + 5 - 3] = (m - √5)(m² + m√5 + 2) = 0 m² + m√5 + 2 > 0, m - √5 = 0, m = a + b = x + 1/x = √5