Solve (a-1)^2-4a^2>0

$$( a - 1 ) ^ { 2 } - 4 a ^ { 2 } > 0$$

Solve for a

$$a\in \left(-1,\frac{1}{3}\right)$$

Steps Using the Quadratic Formula

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

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

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}-2a+1>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -3a^{2}-2a+1 positive. Since -1 is negative, the inequality direction is changed.

$$3a^{2}+2a-1<0$$

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.

$$3a^{2}+2a-1=0$$

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

$$a=\frac{-2±\sqrt{2^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$a=\frac{-2±4}{6}$$

Solve the equation a=\frac{-2±4}{6} when ± is plus and when ± is minus.

$$a=\frac{1}{3} a=-1$$

Rewrite the inequality by using the obtained solutions.

$$3\left(a-\frac{1}{3}\right)\left(a+1\right)<0$$

For the product to be negative, a-\frac{1}{3} and a+1 have to be of the opposite signs. Consider the case when a-\frac{1}{3} is positive and a+1 is negative.

$$a-\frac{1}{3}>0 a+1<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+1 is positive and a-\frac{1}{3} is negative.

$$a+1>0 a-\frac{1}{3}<0$$

The solution satisfying both inequalities is a\in \left(-1,\frac{1}{3}\right).

$$a\in \left(-1,\frac{1}{3}\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-1,\frac{1}{3}\right)$$