Solve (3k-1)^2-4(k-2)(2k+1)geq0

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

Solve for k

$$k\in \mathrm{R}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3k-1\right)^{2}.

$$9k^{2}-6k+1-4\left(k-2\right)\left(2k+1\right)\geq 0$$

Use the distributive property to multiply -4 by k-2.

$$9k^{2}-6k+1+\left(-4k+8\right)\left(2k+1\right)\geq 0$$

Use the distributive property to multiply -4k+8 by 2k+1 and combine like terms.

$$9k^{2}-6k+1-8k^{2}+12k+8\geq 0$$

Combine 9k^{2} and -8k^{2} to get k^{2}.

$$k^{2}-6k+1+12k+8\geq 0$$

Combine -6k and 12k to get 6k.

$$k^{2}+6k+1+8\geq 0$$

Add 1 and 8 to get 9.

$$k^{2}+6k+9\geq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$k^{2}+6k+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 6 for b, and 9 for c in the quadratic formula.

$$k=\frac{-6±\sqrt{6^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$k=\frac{-6±0}{2}$$

Solutions are the same.

$$k=-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(k+3\right)^{2}\geq 0$$

Inequality holds for k\in \mathrm{R}.

$$k\in \mathrm{R}$$