Solve 1/2(a^2-4+1)=1

$$\frac { 1 } { 2 } ( a ^ { 2 } - 4 + 1 ) = 1$$

Solve for a

$$a=\sqrt{5}\approx 2.236067977$$

$$a=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$a^{2}-4+1=1\times 2$$

Multiply 1 and 2 to get 2.

$$a^{2}-4+1=2$$

Add -4 and 1 to get -3.

$$a^{2}-3=2$$

Add 3 to both sides.

$$a^{2}=2+3$$

Add 2 and 3 to get 5.

$$a^{2}=5$$

Take the square root of both sides of the equation.

$$a=\sqrt{5} a=-\sqrt{5}$$

Steps Using the Quadratic Formula

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$a^{2}-4+1=1\times 2$$

Multiply 1 and 2 to get 2.

$$a^{2}-4+1=2$$

Add -4 and 1 to get -3.

$$a^{2}-3=2$$

Subtract 2 from both sides.

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

Subtract 2 from -3 to get -5.

$$a^{2}-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-5\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$a=\frac{0±\sqrt{20}}{2}$$

Take the square root of 20.

$$a=\frac{0±2\sqrt{5}}{2}$$

Now solve the equation a=\frac{0±2\sqrt{5}}{2} when ± is plus.

$$a=\sqrt{5}$$

Now solve the equation a=\frac{0±2\sqrt{5}}{2} when ± is minus.

$$a=-\sqrt{5}$$

The equation is now solved.

$$a=\sqrt{5} a=-\sqrt{5}$$