Solve (4m)^2-3*4(2m^2-2)>0

$$( 4 m ) ^ { 2 } - 3 \times 4 ( 2 m ^ { 2 } - 2 ) > 0$$

Solve for m

$$m\in \left(-\sqrt{3},\sqrt{3}\right)$$

Solution Steps

Expand \left(4m\right)^{2}.

$$4^{2}m^{2}-3\times 4\left(2m^{2}-2\right)>0$$

Calculate 4 to the power of 2 and get 16.

$$16m^{2}-3\times 4\left(2m^{2}-2\right)>0$$

Multiply 3 and 4 to get 12.

$$16m^{2}-12\left(2m^{2}-2\right)>0$$

Use the distributive property to multiply -12 by 2m^{2}-2.

$$16m^{2}-24m^{2}+24>0$$

Combine 16m^{2} and -24m^{2} to get -8m^{2}.

$$-8m^{2}+24>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -8m^{2}+24 positive. Since -1 is negative, the inequality direction is changed.

$$8m^{2}-24<0$$

Add 3 to both sides.

$$m^{2}<3$$

Calculate the square root of 3 and get \sqrt{3}. Rewrite 3 as \left(\sqrt{3}\right)^{2}.

$$m^{2}<\left(\sqrt{3}\right)^{2}$$

Inequality holds for |m|<\sqrt{3}.

$$|m|<\sqrt{3}$$

Rewrite |m|<\sqrt{3} as m\in \left(-\sqrt{3},\sqrt{3}\right).

$$m\in \left(-\sqrt{3},\sqrt{3}\right)$$