Q1.If α (alpha) and β (beta) be two roots of the equation x² – 64x + 256 = 0. Then the value of –

(a.) (α³/β⁵)¹/⁸ + (β

2 thoughts on “Q1.If α (alpha) and β (beta) be two roots of the equation x² – 64x + 256 = 0. Then the value of –<br /><br /> (a.) (α³/β⁵)¹/⁸ + (β”

1. Step by step shortcut explanation

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2. Explanation:

   ✬ Values = 2 & 8 Respectively ✬

   Step-by-step explanation:

   Given:

   Alpha and beta are the two roots of equation.

   Equation is x² – 64x + 256 = 0

   To Find:

   Value of

   (α³/β⁵)¹/⁸ + (β³/α⁵)¹/⁸

   (256α/α² + 256) + (256β/β² + 256)

   Solution: Basic formulae and concepts to be used here

   α + β = –b/a

   αβ = c/a

   m⁵ × m³ = m(³ + ⁵) ← If bases are same then powers will be added.

   m² × n² = (mn)² ← If bases are different and powers are same.

   Let’s solve the first one –

   ➟ (α³/β⁵)¹/⁸ + (β³/α⁵)¹/⁸

   ➟ α⅜ / β⅝ + β⅜ / α⅝ { taking LCM }

   ➟ α⅜ × β⅝ + β⅝ × β⅜ / β⅝ × α⅝

   ➟ α¹ + β¹ / αβ⅝

   ➟ α + β / αβ⅝

   ➟ –b/a / (c/a)⅝

   ➟ –(–64) / (256)⅝

   ➟ 64 / {(16)²}⅝

   ➟ 64 / [{(2)⁴}²]⅝

   ➟ 64 / 2⁵

   ➟ 64 / 32 = 2

   Hence, the value of (α³/β⁵)¹/⁸ + (β³/α⁵)¹/⁸ is 2.

   _________________________

   Let’s move to the second question.

   (256α/α² + 256) + (256β/β² + 256)

   ➮ Given equation is x² – 64x + 256 = 0

   Putting the value α in the place or x in above equation.

   We got α² – 64α + 256 = 0

   ➮ α² – 64α + 256 = 0

   ➮ α² + 256 = 64α

   Multiplying both sides by 4.

   ➮ 4 × (α² + 256) = 4 × 64α

   ➮ 4 x (α² + 256) = 256α

   ➮ 4 = 256α / α² + 256ㅤㅤㅤㅤㅤ(eqⁿ i )

   Now again

   Putting the value β in the place or x in above equation.

   We got β² – 64β + 256 = 0

   ➯ β² – 64β + 256 = 0

   ➯ β² + 256 = 64β

   Multiplying both sides by 4.

   ➯ 4 × (β² + 256) = 4 × 64β

   ➯ 4 = 256β / β² + 256ㅤㅤㅤㅤㅤ(eqⁿ ii )

   Let’s put the values of both equations in the second question.

   ⟹ (256α/α² + 256) + (256β/β² + 256)

   ⟹ 4 + 4

   =8

   Hence, the value of (256α/α² + 256) + (256β/β² + 256) is 8.

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