Solve 4(k+1)^2-4(k+1)(k-2)geq0

$$4 ( k + 1 ) ^ { 2 } - 4 ( k + 1 ) ( k - 2 ) \geq 0$$

Solve for k

$$k\geq -1$$

Solution Steps

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

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

Use the distributive property to multiply 4 by k^{2}+2k+1.

$$4k^{2}+8k+4-4\left(k+1\right)\left(k-2\right)\geq 0$$

Use the distributive property to multiply -4 by k+1.

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

Use the distributive property to multiply -4k-4 by k-2 and combine like terms.

$$4k^{2}+8k+4-4k^{2}+4k+8\geq 0$$

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

$$8k+4+4k+8\geq 0$$

Combine 8k and 4k to get 12k.

$$12k+4+8\geq 0$$

Add 4 and 8 to get 12.

$$12k+12\geq 0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$12k\geq -12$$

Divide both sides by 12. Since 12 is positive, the inequality direction remains the same.

$$k\geq \frac{-12}{12}$$

Divide -12 by 12 to get -1.

$$k\geq -1$$