Solve [-8k(2k-1)]^2-4*(3+4k^2)*(16k^2-16k-8)>0

$$[ - 8 k ( 2 k - 1 ) ] ^ { 2 } - 4 \cdot ( 3 + 4 k ^ { 2 } ) \cdot ( 16 k ^ { 2 } - 16 k - 8 ) > 0$$

Solve for k

$$k>-\frac{1}{2}$$

Solution Steps

Use the distributive property to multiply -8k by 2k-1.

$$\left(-16k^{2}+8k\right)^{2}-4\left(3+4k^{2}\right)\left(16k^{2}-16k-8\right)>0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-16k^{2}+8k\right)^{2}.

$$256\left(k^{2}\right)^{2}-256k^{2}k+64k^{2}-4\left(3+4k^{2}\right)\left(16k^{2}-16k-8\right)>0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$256k^{4}-256k^{2}k+64k^{2}-4\left(3+4k^{2}\right)\left(16k^{2}-16k-8\right)>0$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$256k^{4}-256k^{3}+64k^{2}-4\left(3+4k^{2}\right)\left(16k^{2}-16k-8\right)>0$$

Use the distributive property to multiply -4 by 3+4k^{2}.

$$256k^{4}-256k^{3}+64k^{2}+\left(-12-16k^{2}\right)\left(16k^{2}-16k-8\right)>0$$

Use the distributive property to multiply -12-16k^{2} by 16k^{2}-16k-8 and combine like terms.

$$256k^{4}-256k^{3}+64k^{2}-64k^{2}+192k+96-256k^{4}+256k^{3}>0$$

Combine 64k^{2} and -64k^{2} to get 0.

$$256k^{4}-256k^{3}+192k+96-256k^{4}+256k^{3}>0$$

Combine 256k^{4} and -256k^{4} to get 0.

$$-256k^{3}+192k+96+256k^{3}>0$$

Combine -256k^{3} and 256k^{3} to get 0.

$$192k+96>0$$

Subtract 96 from both sides. Anything subtracted from zero gives its negation.

$$192k>-96$$

Divide both sides by 192. Since 192 is positive, the inequality direction remains the same.

$$k>\frac{-96}{192}$$

Reduce the fraction \frac{-96}{192} to lowest terms by extracting and canceling out 96.

$$k>-\frac{1}{2}$$