Math, asked by rithu1404, 7 months ago

IF
x ^2 + 1/ x^2 = 7 Find the value of x + 1/x​

Answers

Answered by krishna275930
0

sjsjsjshdfjicuahwhwhejdidid

Answered by dhamija123
4

hey mate here is ur answer

x^2+(1/x^2)=7

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitute

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5take the square root of both sides

x^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=7x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5x^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5take the square root of both sidesx^2+(1/x^2)=7x-(1/x)=??square x-(1/x)x^2-2+(1/x^2)substitutex^2+(1/x^2)=77-2=5(x-[1/x])^2=5take the square root of both sidesx-(1/x)=√5

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