Solve (2k-1)^2-4(k^2+3)>0

$$( 2 k - 1 ) ^ { 2 } - 4 ( k ^ { 2 } + 3 ) > 0$$

Solve for k

$$k<-\frac{11}{4}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2k-1\right)^{2}.

$$4k^{2}-4k+1-4\left(k^{2}+3\right)>0$$

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

$$4k^{2}-4k+1-4k^{2}-12>0$$

Combine 4k^{2} and -4k^{2} to get 0.

$$-4k+1-12>0$$

Subtract 12 from 1 to get -11.

$$-4k-11>0$$

Add 11 to both sides. Anything plus zero gives itself.

$$-4k>11$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$k<-\frac{11}{4}$$