Question

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 + (β³/α⁵)¹/⁸

(b.) (256α/α² + 256) + (256β/β² + 256)

JEE MAINS - 2020​

Answer 1

✬ 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

1. (α³/β⁵)¹/⁸ + (β³/α⁵)¹/⁸
2. (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.

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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.

$\implies{\rm }$ (256α/α² + 256) + (256β/β² + 256)

$\implies{\rm }$ 4 + 4

$\implies{\rm }$ 8

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

Answer 2

Given:-

• α and β be two roots of the equation $x² – 64x + 256 = 0.$

To Find:-

• Value of $(\dfrac{α³}{β³})^{⅛}+(\dfrac{β³}{α³})^{⅛}$

• $(\dfrac{256α}{α² + 256}) + (\dfrac{256β}{β² + 256})$

Formula used:-

• α + β = $\dfrac{-b}{a}$

• αβ = $\dfrac{c}{a}$

Solution:-

$⟹(\dfrac{α³}{β³})^{⅛}+(\dfrac{β³}{α³})^{⅛}$

$⟹(\dfrac{α^{⅜}}{β^{⅝}})+(\dfrac{β^{⅜}}{α^{⅝}})$

$⟹\dfrac{α^{⅜}×α^{⅝}+β^{⅜}×β^{⅝}}{(αβ)^{⅝}}$

$⟹\dfrac{α^{\frac{8}{8}}+β^{\frac{8}{8}}}{(αβ)^{⅝}}$

$⟹\dfrac{α¹+β¹}{(αβ)^{⅝}}$

$⟹\dfrac{α + β}{αβ^{⅝}}$

$⟹\dfrac{\dfrac{-b}{a}}{(\dfrac{c}{a})^{⅝}}$

$⟹\dfrac{–(–64)}{(256)^{⅝}}$

$⟹\dfrac{64}{(16)^{2×\frac{5}{8}}}$

$⟹\dfrac{64}{(4)^{2×\frac{5}{4}}}$

$⟹\dfrac{64}{(2)^{2×\frac{5}{2}}}$

$⟹\dfrac{64}{2⁵}$

$⟹\dfrac{64}{32}$

$⟹\:2$

Hence, Value of $(\dfrac{α³}{β³})^{⅛}+(\dfrac{β³}{α³})^{⅛}$ is $\:2.$

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Now, Solving second part:-

$⟹\:(\dfrac{256α}{α² + 256}) + (\dfrac{256β}{β² + 256})$

As we know, α and β be two roots of the equation $x² – 64x + 256 = 0.$

So, by substituting the value of x by α.

$⟹\:α²\: –\: 64α + 256 = 0$

$⟹\:α² \:+\: 256 = 64α$

We can't simply further so, we will multiply 4 both the sides to simplify further.

$⟹\:4(α² + 256) = (64α)×4$

$⟹\:4(α² + 256) = 256α$

$⟹\:\dfrac{256α}{α² + 256}=4$ ____{1}

Now, we will substitute x with β.

$⟹\:β²\: – \:64β + 256 = 0$

$⟹\:β² \: +\: 256 = 64β$

We can't simply further so, we will multiply 4 both the sides to simplify further.

$⟹\:4(β² + 256) = (64β)×4$

$⟹\:4(β² + 256) = 256β$

$⟹\:\dfrac{256β}{β² + 256}=4$ ____{2}

Putting {1} and {2} in $(\dfrac{256α}{α² + 256}) + (\dfrac{256β}{β² + 256})$

$⟹\:(\dfrac{256α}{α² + 256}) + (\dfrac{256β}{β² + 256})$

$⟹\:4+4$

$⟹\:8$

Hence, the value of ${\bold{(\dfrac{256α}{α² + 256}) + (\dfrac{256β}{β² + 256})}}$ is 8.

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