Math, asked by ashokkumar550h, 6 hours ago

find the value of x:
1. \sqrt[3]{2x - 1}  - 3 = 0 \\ 2. \sqrt[5]{3x + 2}  - 2 = 0

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Answers

Answered by mathdude500
3

\large\underline{\sf{Solution-i}}

\rm :\longmapsto\: \sqrt[3]{2x - 1} - 3 = 0

can be rewritten as

\rm :\longmapsto\: \sqrt[3]{2x - 1} = 3

On Cubing both sides, we get

\rm :\longmapsto\:2x - 1 =  {(3)}^{3}

\rm :\longmapsto\:2x - 1 =27

\rm :\longmapsto\:2x =27  + 1

\rm :\longmapsto\:2x =28

\bf\implies \:x = 14

Verification :-

Consider LHS

\rm :\longmapsto\: \sqrt[3]{2x - 1} - 3

On Substituting x = 14, we get

\rm \:  =  \:  \:  \sqrt[3]{2 \times 14 - 1} - 3

\rm  \:  =  \:  \:  \sqrt[3]{28 - 1} - 3

\rm  \:  =  \:  \:  \sqrt[3]{27} - 3

\rm  \:  =  \:  \:  \sqrt[3]{3 \times 3 \times 3} - 3

\rm  \:  =  \:  \:  3  - 3

\rm \:  =  \:  \:0

Hence, Verified

 \red{\large\underline{\sf{Solution-ii}}}

\rm :\longmapsto\: \sqrt[5]{3x + 2} - 2 = 0

can be rewritten as

\rm :\longmapsto\: \sqrt[5]{3x + 2} = 2

On taking 5 power on both sides, we get

\rm :\longmapsto\:3x + 2 =  {(2)}^{5}

\rm :\longmapsto\:3x + 2 = 32

\rm :\longmapsto\:3x = 32  - 2

\rm :\longmapsto\:3x = 30

\bf\implies \:x = 10

Verification :-

Consider LHS

\rm :\longmapsto\: \sqrt[5]{3x + 2} - 2

On substituting x = 10, we get

\rm \:  =  \:  \: \sqrt[5]{3 \times 10 + 2}  - 2

\rm \:  =  \:  \: \sqrt[5]{30 + 2}  - 2

\rm \:  =  \:  \: \sqrt[5]{32}  - 2

\rm \:  =  \:  \: \sqrt[5]{2 \times 2 \times 2 \times 2 \times 2}  - 2

\rm \:  =  \:  \: 2 - 2

\rm \:  =  \:  \:0

Hence, Verified

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