Solve 4x^2+2x+2x+1>-16x^2

$$4 x ^ { 2 } + 2 x + 2 x + 1 > - 16 x ^ { 2 }$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Combine 2x and 2x to get 4x.

$$4x^{2}+4x+1>-16x^{2}$$

Add 16x^{2} to both sides.

$$4x^{2}+4x+1+16x^{2}>0$$

Combine 4x^{2} and 16x^{2} to get 20x^{2}.

$$20x^{2}+4x+1>0$$

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.

$$20x^{2}+4x+1=0$$

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 20 for a, 4 for b, and 1 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 20\times 1}}{2\times 20}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-64}}{40}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 20x^{2}+4x+1 has the same sign for any x. To determine the sign, calculate the value of the expression for x=0.

$$20\times 0^{2}+4\times 0+1=1$$

The value of the expression 20x^{2}+4x+1 is always positive. Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$