Solve the following equations : (i) 3x$^{2}$ - 4 $\sqrt{3x^{2} - 4x +1}$ = 4x - 4. (ii) x$^{2}$ - $\sqrt{2x^{2} -8x + 12}$ = 4x + 6. find any one value of x

Checking the options against the second equation x^2 - sqrt(2x^2 - 8x + 12) = 4x + 6 reveals that x = -2 is a solution. Substituting x = -2 yields (-2)^2 - sqrt(2(-2)^2 - 8(-2) + 12) = 4(-2) + 6, which simplifies to 4 - sqrt(16 + 16 + 12) = -8 + 6. This results in 4 - 6 = -2, which confirms the value.

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