RADICAL Math Problem Stumps 95% of People!

[@RashmiRay-c1y]

E=[44√3 + 76]^1/5. So, E^5= 44√3 + 76 and let F^5= 44√3 - 76. Then (EF)^5 = 32 = 2^5. So, EF=2 and E^5-F^5=152. Let t = E-F. Then, t^5 + 10 t^3 +20 t = 152. So, t =2. With E-F=2 and EF=2 and with E positive, we get E = √3 + 1.

[@ashokdubey8415]

5:00 How did you conclude x>2 ?

[@Quest3669]

?^5= 77+44√ 3 hence ?= √3+1 >0

[@潘博宇-k4l]

E=1+(3)^(1/2).

[@gregevgeni1864]

1+√3

[@walterwen2975]

RADICAL Math Problem: ⁵√[(√16 + √48)/(√49 - √48)] =? (√16 + √48)/(√49 - √48) = 4(1 + √3)/(7 - 4√3) > 0 (√16 + √48)/(√49 - √48) = [4(1 + √3)(7 + 4√3)]/[(7 - 4√3)(7 + 4√3)] = [(1 + √3)(4)(7 + 4√3)]/(49 - 48) = (1 + √3)(28 + 16√3) (1 + √3)⁴ = [(1 + √3)²]² = (4 + 2√3)² = 4(2 + √3)² = 4(7 + 4√3) = 28 + 16√3 (√16 + √48)/(√49 - √48) = (1 + √3)(28 + 16√3) = (1 + √3)[(1 + √3)⁴] = (1 + √3)⁵ ⁵√[(√16 + √48)/(√49 - √48)] = ⁵√[(1 + √3)⁵] = 1 + √3 Final answer: (√16 + √48)/(√49 - √48) = 1 + √3

[@Fjfurufjdfjd]

χ=[76+44(3)^(1/2)]^(1/5)>0.καταληγω στο συστημα χ^5+ψ^5=152 χψ=-1 οπουψ=[76-44(3)^(1/2)]^(1/5)0 Για το α εχω α^5+10α^3+20α-152=0 (α-2)(α^4+2α^3+14α^2+28α+76)=0. Η δευτερη παρενθεση δεν εχει θετικη ριζα. Αρα α=2. χ+ψ=α>0 αρα χ=1+(3)^(1/2)

[@kaushalagrawal6258]

THIS LOOKS SO COOL

[@Fjfurufjdfjd]

Ευχαριστω.

[@ABHISHEKKUMAR-01024]

We have, (√16 + √48) / (√49 - √48) = (4 + 4√3) / (7 - 4√3) = 4(1 + √3) / (7 - 4√3) = 4(1 + √3)(7 + 4√3) / (7 - 4√3)(7 + 4√3) = 4 { (7 + 4√3) + √3(7 + 4√3) } / { 7² - (4√3)² } = 4 (7 + 4√3 + 7√3 + 12) / (49 - 48) = 4 (19 + 11√3) = 76 + 44√3 Let ⁵√{ (√16 + √48) / (√49 - √48) } = a Then, ⁵√(76 + 44√3) = a Also, 76 - 44√3 = 4 (19 - 11√3) = 4 (19 - 11√3) (19 + 11√3) / (19 + 11√3) = 4 { 19² - (11√3)² } / (19 + 11√3) = 4 { 361 - (121 × 3) } / (19 + 11√3) = 4 (361 - 363) / (19 + 11√3) = - 8 / (19 + 11√3) Therefore, 76 - 44√3 < 0 Now, let ⁵√(44√3 - 76) = b, b > 0 Now, a⁵ = 76 + 44√3 and b⁵ = 44√3 - 76 => a⁵ - b⁵ = 152 Also, ab = { ⁵√(76 + 44√3) } { ⁵√(44√3 - 76) } = [⁵√{4(19 + 11√3)}] [⁵√{4(11√3 - 19)}] = ⁵√{ 16(19 + 11√3)(11√3 - 19) } = ⁵√[ 16 {(11√3)² - 19²} ] = ⁵√{16 (363 - 361) } = ⁵√(16 × 2) = ⁵√32 = 2 Now, (a - b)⁵ = a⁵ - 5a⁴b + 10a³b² - 10a²b³ + 5ab⁴ - b⁵ = (a⁵ - b⁵) - (5a⁴b - 5ab⁴) + (10a³b² - 10a²b³) = (a⁵ - b⁵) - 5ab(a³ - b³) + 10a²b²(a - b) Now, (a - b)³ = a³ - 3a²b + 3ab² - b³ = (a³ - b³) - (3a²b - 3ab²) = (a³ - b³) - 3ab(a - b) Therefore, (a³ - b³) = (a - b)³ + 3ab(a - b) Substituting, (a - b)⁵ = (a⁵ - b⁵) - 5ab {(a - b)³ + 3ab(a - b) } + 10a²b²(a - b) = (a⁵ - b⁵) - 5ab(a - b)³ - 15a²b²(a - b) + 10a²b²(a - b) = (a⁵ - b⁵) - 5ab(a - b)³ - 5a²b²(a - b) = (a⁵ - b⁵) - 5ab(a - b)³ - 5(ab)²(a - b) Substituting a⁵ - b⁵ = 152 and ab = 2, (a - b)⁵ = 152 - 5(2)(a - b)³ - 5(2²)(a - b) or, (a - b)⁵ = 152 - 10(a - b)³ - 5(4)(a - b) or, (a - b)⁵ = 152 - 10(a - b)³ - 20(a - b) or, (a - b)⁵ + 10(a - b)³ + 20(a - b) - 152 = 0 or, α⁵ + 10α³ + 20α - 152 = 0, where α = a - b We have, 2⁵ + 10(2³) + 20(2) - 152 = 32 + 10(8) + 40 - 152 = 32 + 80 + 40 - 152 = 0 Therefore, α - 2 is one of the factors. Then, α⁵ + 10α³ + 20α - 152 = 0 or, α⁵ - 2α⁴ + 2α⁴ - 4α³ + 14α³ - 28α² + 28α² - 56α + 76α - 152 = 0 or, (α⁵ - 2α⁴) + (2α⁴ - 4α³) + (14α³ - 28α²) + (28α² - 56α) + (76α - 152) = 0 or, α⁴(α - 2) + 2α³((α - 2) + 14α²(α - 2) + 28α(α - 2) + 76(α - 2) = 0 or, (α - 2) (α⁴ + 2α³ + 14α² + 28α + 76) = 0 or, α - 2 = 0, α⁴ + 2α³ + 14α² + 28α + 76 = 0 or, α = 2, α⁴ + 2α³ + 14α² + 28α + 76 = 0 Let us focus our attention on the root α = 2 Then a - b = 2 Also ab = 2 Now, (a + b)² = (a - b)² + 4ab = 2² + 4(2) = 4 + 8 = 12 Therefore, a + b = 2√3 (We discard the - ve sign as a > 0 and b > 0) Then, 2a = (a + b) + (a - b) = 2√3 + 2 = 2(√3 + 1) or, a = √3 + 1 or, ⁵√{ (√16 + √48) / (√49 - √48) } = √3 + 1 The required answer.