Solve (x+1)^2>4x

$$( x + 1 ) ^ { 2 } > 4 x$$

Solve for x

$$x\neq 1$$

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}.

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

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

$$x^{2}-2x+1>0$$

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.

$$x^{2}-2x+1=0$$

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 1 for a, -2 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{2±0}{2}$$

Solutions are the same.

$$x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-1\right)^{2}>0$$

Inequality holds for x\neq 1.

$$x\neq 1$$