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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 27 July · Morning Shift · Mathematics · Quadratic Equation

Let α, β be two roots of the equation x ^ 2 + (20) ^ 1 / 4 x + (5)...

Let α,β\alpha, \beta be two roots of the equation x2+(20)1/4x+(5)1/2=0{x}^{2} + {\left(20\right)}^{1 / 4}x + {\left(5\right)}^{1 / 2} = 0. Then α8+β8{\alpha}^{8} + {\beta}^{8} is equal to

Options

  1. A1010
  2. B100100
  3. C5050
  4. D160160

Correct answer

C. 5050

Step-by-step solution

Given,

x2+(20)1/4x+(5)1/2=0{x}^{2} + {\left(20\right)}^{1 / 4}x + {\left(5\right)}^{1 / 2} = 0

⇒x2+(5)=−x(20)14⇒ {x}^{2} + \left(\sqrt{5}\right) = - x\left(20\right)\dfrac{1}{4}

⇒(x2+5)2=20x2⇒ {\left({x}^{2} + \sqrt{5}\right)}^{2} = \sqrt{20}{x}^{2}

⇒x4+5+2×x2×5=20x2⇒ {x}^{4} + 5 + 2 \times {x}^{2} \times \sqrt{5} = \sqrt{20}{x}^{2}

⇒x4=−5⇒x8=25⇒ {x}^{4} = - 5 ⇒ {x}^{8} = 25

So, α8+β8=25+25=50{\alpha}^{8} + {\beta}^{8} = 25 + 25 = 50

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