Solve left(x+1right)^2+3geqx

$${ \left(x+1 \right) }^{ 2 } +3 \geq x$$

Solve for x

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

Steps Using the Quadratic Formula

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

$$x^{2}+2x+1+3\geq x$$

Add 1 and 3 to get 4.

$$x^{2}+2x+4\geq x$$

Subtract x from both sides.

$$x^{2}+2x+4-x\geq 0$$

Combine 2x and -x to get x.

$$x^{2}+x+4\geq 0$$

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.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-15}}{2}$$

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

$$0^{2}+0+4=4$$

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

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