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Question
`sqrt2x^3-3x-2sqrt2=0`
Solution
`sqrt2x^3-3x-2sqrt2=0`
`⇒2x^2-3x-2sqrt2=0` (Multiplying both sides by `sqrt2`)
⇒`2x^2-3sqrt2x=4`
⇒`(sqrt2x)^2-2xxsqrt2x xx3/2+(3/2)^2=4+(3/2)^2` [Adding `(3/2)^2` on both sides]
⇒`(sqrt2x-3/2)^2=4+9/4=25/4=(5/2)^2`
⇒`sqrt2x-3/2=+-5/2` (Taking square root on both sides)
⇒`sqrt2x-3/2=5/2 or sqrt2x-3/2=-5/2`
⇒`sqrt2x=5/2+3/2=8/2=4 or sqrt2x=-5/2+3/2=-2/2=-1`
⇒`x=4/sqrt2=2sqrt2 or x=-1/sqrt2=-sqrt2/2`
Hence, `2sqrt2` and `-sqrt2/2` are the roots of the given equation.
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