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[tex]y =x^2(2+4)\\\\\implies 2= 6x^2 \\\\\implies x^2 = \dfrac 26 = \dfrac 13\\\\\implies x = \pm\dfrac 1{\sqrt 3}[/tex]
[tex]\text{Hence}~ x = \dfrac 1{\sqrt 3} ~\text{or}~ ~ x = -\dfrac 1{\sqrt 3}[/tex]