Solve (2k+3)^2+k^2>0

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

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(2k+3\right)^{2}.

$$4k^{2}+12k+9+k^{2}>0$$

Combine 4k^{2} and k^{2} to get 5k^{2}.

$$5k^{2}+12k+9>0$$

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.

$$5k^{2}+12k+9=0$$

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 5 for a, 12 for b, and 9 for c in the quadratic formula.

$$k=\frac{-12±\sqrt{12^{2}-4\times 5\times 9}}{2\times 5}$$

Do the calculations.

$$k=\frac{-12±\sqrt{-36}}{10}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 5k^{2}+12k+9 has the same sign for any k. To determine the sign, calculate the value of the expression for k=0.

$$5\times 0^{2}+12\times 0+9=9$$

The value of the expression 5k^{2}+12k+9 is always positive. Inequality holds for k\in \mathrm{R}.

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