Question

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Answer 1

Question

If ³√a + 64/³√a = 20 then the value of q can be :

(A) 4 & 32

(B) 8 & 4

(C) 16 & 8

(D) 4 & 16

(E) None of these

Solution

$\sqrt[3]{a} + \dfrac{64}{ \sqrt[3]{a} } = 20$

Substituting, x = ³√a in the equation

$\implies x + \dfrac{64}{x} = 20$

Taking LCM

$\implies \dfrac{ {x}^{2} + 64 }{x} = 20$

$\implies {x}^{2} + 64 = 20x$

$\implies {x}^{2} - 20x+ 64 = 0$

Splitting the middle term

$\implies {x}^{2} -16x - 4x+ 64 = 0$

$\implies x(x - 16) - 4(x - 16)= 0$

$\implies (x - 4)(x - 16)= 0$

$\implies x - 4 = 0 \quad or \quad x - 16 = 0$

$\implies x = 4 \quad or \quad x = 16$

$\implies \sqrt[3]{a} = 4 \quad or \quad \sqrt[3]{a} = 16$

$\implies a = {4}^{3} \quad or \quad a = {16}^{3}$

Hence, the answer is (E) None of these.

Answer 2

Question :-

$\sf{if \: \sqrt[3]{a} + \frac{64}{ \sqrt[3]{a} } = 20 \: then \: find \: value \: of \: a}$

(A) 4 and 32

(B) 8 and 4

(C) 16 and 8

( D) 4 and 16

(E) None of these

Answer :-

→ Option ( E ) is correct (None of these )

To Find :-

→ Value of a

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Step - by - step explanation :-

According to the question,

$\rightarrow \red{\sf{\sqrt[3]{a} + \frac{64}{ \sqrt[3]{a} }} = 20 \: } \\ \\ \sf{assume \: \: \orange{y \: = \green{ \sqrt[3]{a} }} }\\ \\ \rightarrow \sf{ y + \red{\frac{64}{y} } = \blue{ 20}} \\ \\ \rightarrow \sf{ \orange{ \frac{ {y}^{2} + 64 }{y} }= 20 } \\ \\ \rightarrow \sf{ \red{ {y}^{2} }+ 64 = \green{ 20y}} \\ \\ \rightarrow \sf{ {y}^{2} \blue{ - 20y + }64} = 0\\ \\ \text{ \red{solve }\: by \: \pink{factorisation }\: \orange{method\: }} \\ \\ \rightarrow \sf{ \green{{y}^{2} - 16y }- 4y + \red{64= 0}} \\ \\ \rightarrow \sf{ y( \orange{y - 16}) \red{ - 4}( \orange{y - 16}) = \green{0}} \\ \\ \rightarrow \sf{ \red{ (y - 16)} \green{(y - 4) }= 0} \\ \\ \star \: \text{case (1)} \\ \\ \implies \sf{ \red{y - 16 }= 0} \\ \\ \implies \sf{ \green{y \: = 16}} \: \: \: \: \: \: \because \sf{ y \: = \blue{\sqrt[3]{a} }} \\ \\ \implies \sf{ \red{ \sqrt[3]{a} }= 16} \\ \\ \implies \boxed{ \sf{ \orange{a = {(16)}^{3}} }} \\ \: \\ \star \text{ case (2)} \\ \\ \implies \sf{ \red{ y - 4 }= 0} \\ \\ \implies \sf{ \green{ y = 4}} \: \: \: \: \because \: y \: = \pink{\sqrt[3]{a} } \\ \\ \rightarrow \sf{ \green{\sqrt[3]{a}} = 4} \\ \\ \implies \boxed{ \sf{ \red{a \: = {(4)}^{3} }}}$

Option ( E) is correct (None of these ).

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