Question

$\mathcal{\gray{\underbrace{\orange{Q}\green{u}\orange{E}\green{s}\orange{T}\green{i}\orange{O}\green{n}\:}}}$

️ JEE-MAIN - 06/09/2020
⚡ If $\rm{\alpha\:,\:\beta}$ are the roots of the equation “x² - 64x + 256 = 0” . Then find the value of $\rm\bold{\left\{\:\dfrac{{\alpha}^3}{{\beta}^5}\right\}^{\dfrac{1}{8}}\:+\:\left\{\:\dfrac{{\beta}^3}{{\alpha}^5}\right\}^{\dfrac{1}{8}}\:}$ .
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✍️ Plz Don't Spam .
✍️ Answer With Proper EXPLANATION .​

Answer 1

Solution :-

Comparing the given Quadratic Equation x² - 64x + 256 = 0 with ax² + bx + c = 0 , we get :-

• a = 1
• b = (-64)
• c = 256

As, we have given that, ɑ and β are are the roots of equation .

So,

→ sum of roots = (ɑ + β) = (-b/a) = -(64)/1 = 64 .

→ Product of roots = ɑ * β = (c/a) = (256/1) = 256 .

Now,

we have to Find :-

→ (ɑ³/β⁵)^(1/8) + (β³/ɑ⁵)^(1/8)

→ {(ɑ³)^(1/8)}/{(β⁵)^(1/8)} + {(β³)^(1/8)}/{(ɑ⁵)^(1/8)}

→ {ɑ^(3/8) / β^(5/8)} + {β^(3/8) / ɑ^(5/8)}

Taking LCM Now,

→ {ɑ^(3/8) * ɑ^(5/8) + β^(3/8) * β^(5/8)} / { ɑ^(5/8) * β^(5/8)}

→ {ɑ^(3/8 + 5/8) + β^(3/8 + 5/8)} / {(ɑβ)^(5/8)}

→ (ɑ + β) / {(ɑβ)^(5/8)

Putting values Now,

→ 64 / (256)^(5/8)

→ 64 / {(2)^8}^(5/8)

→ 64 / (2)⁵

→ 64 / 32

→ 2 (Ans.)

Therefore, Correct Answer will be 2.

Answer 2

Question :

If α , β are roots of the equation x² - 64x + 256 = 0 , then find the value of $\sf \Bigg\{\dfrac{\alpha^3}{\beta^5}\Bigg\}^{\dfrac{1}{8}}+\Bigg\{\dfrac{\beta^3}{\alpha^5}\Bigg\}^{\dfrac{1}{8}}$

Solution :

Let's simplify required value ,

$:\to \sf \Bigg\{\dfrac{\alpha^3}{\beta^5}\Bigg\}^{\dfrac{1}{8}}+\Bigg\{\dfrac{\beta^3}{\alpha^5}\Bigg\}^{\dfrac{1}{8}}$

$:\to \sf \dfrac{\alpha^{\frac{3}{8}}}{\beta^{\frac{5}{8}}}+\dfrac{\beta^{\frac{3}{8}}}{\alpha^{\frac{5}{8}}}$

$:\to \sf \dfrac{\alpha^{\frac{3}{8}}.\alpha^{\frac{5}{8}}+\beta^{\frac{3}{8}}.\beta^{\frac{5}{8}}}{\alpha^{\frac{5}{8}}.\beta^{\frac{5}{8}}}$

$\sf \bullet\ \; a^m.a^n=a^{m+n}$

$:\to \sf \dfrac{\alpha^{\frac{8}{8}}+\beta^{\frac{8}{8}}}{(\alpha.\beta)^{\frac{5}{8}}}$

$:\leadsto \sf \dfrac{\alpha+\beta}{(\alpha.\beta)^{\frac{5}{8}}}\ \; \pink{\bigstar}$

★══════════════════════★

Compare given equation x² - 64x + 256 with ax² + bx + c , we get ,

• a = 1  ,  b = -64  ,  c = 256

So ,

➠ Sum of zeroes = $\sf -\dfrac{b}{a}$

➠ α + β = $\sf -\dfrac{(-64)}{1}$

➠ α + β = 64  ... (1)

➳ Product of zeroes = $\sf \dfrac{c}{a}$

➳ αβ = $\sf \dfrac{256}{1}$

➳ αβ = 256 ... (2)

★══════════════════════★

Sub. (1) and (2) in our required value ,

$:\implies \sf \Bigg\{\dfrac{\alpha^3}{\beta^5}\Bigg\}^{\dfrac{1}{8}}+\Bigg\{\dfrac{\beta^3}{\alpha^5}\Bigg\}^{\dfrac{1}{8}}$

$:\implies \sf \dfrac{\alpha+\beta}{(\alpha\beta)^{\frac{5}{8}}}$

$:\implies \sf \dfrac{64}{(256)^{\frac{5}{8}}}$

$:\implies \sf \dfrac{64}{\big(2^8\big)^{\frac{5}{8}}}$

$:\implies \sf \dfrac{64}{2^5}$

$:\implies \sf \dfrac{64}{32}$

$:\implies \sf 2\ \; \green{\bigstar}$

Required value is 2