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solve for x sqrt(x^2+x+1) + sqrt(x^2-x+1) >= sqrt(2x^2+6x+2)

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user8121921 | Student, Grade 10 | (Level 1) eNoter

Posted March 31, 2013 at 2:11 PM via web

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solve for x

sqrt(x^2+x+1) + sqrt(x^2-x+1) >= sqrt(2x^2+6x+2)

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted March 31, 2013 at 2:34 PM (Answer #1)

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`sqrt(x^2+x+1)+sqrt(x^2-x+1)>=sqrt(2x^2+6x+2)`

Squaring both side

`x^2+x+1+x^2-x+1+2sqrt((x^2+1)^2-x^2)>=2x^2+6x+2`

simplify ,we have

`sqrt((x^2+)^2-x^2)>=3x`

squaring both side again

`(x^2+1)^2-x^2>=9x^2`

`x^4+1+2x^2-x^2-9x^2>=0`

`x^4-8x^2+1>=0`

`(x^2-4)^2>=(+-sqrt(15))^2`

taking positive square roots

`x^2-4>=+-sqrt(15)`

`x^2>=4+-sqrt(15)`

`x>=+-sqrt(4+-sqrt(15))`

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