Solve b^2=(sqrt{5})^2-(2sqrt{5})^2

$$b ^ { 2 } = ( \sqrt { 5 } ) ^ { 2 } - ( 2 \sqrt { 5 } ) ^ { 2 }$$

Solve for b

$$b=-\sqrt{15}i\approx -0-3.872983346i$$

$$b=\sqrt{15}i\approx 3.872983346i$$

Steps by Finding Square Root

The square of \sqrt{5} is 5.

$$b^{2}=5-\left(2\sqrt{5}\right)^{2}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$b^{2}=5-2^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$b^{2}=5-4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$b^{2}=5-4\times 5$$

Multiply 4 and 5 to get 20.

$$b^{2}=5-20$$

Subtract 20 from 5 to get -15.

$$b^{2}=-15$$

The equation is now solved.

$$b=\sqrt{15}i b=-\sqrt{15}i$$

Steps Using the Quadratic Formula

The square of \sqrt{5} is 5.

$$b^{2}=5-\left(2\sqrt{5}\right)^{2}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$b^{2}=5-2^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$b^{2}=5-4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$b^{2}=5-4\times 5$$

Multiply 4 and 5 to get 20.

$$b^{2}=5-20$$

Subtract 20 from 5 to get -15.

$$b^{2}=-15$$

Add 15 to both sides.

$$b^{2}+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 15}}{2}$$

Square 0.

$$b=\frac{0±\sqrt{-4\times 15}}{2}$$

Multiply -4 times 15.

$$b=\frac{0±\sqrt{-60}}{2}$$

Take the square root of -60.

$$b=\frac{0±2\sqrt{15}i}{2}$$

Now solve the equation b=\frac{0±2\sqrt{15}i}{2} when ± is plus.

$$b=\sqrt{15}i$$

Now solve the equation b=\frac{0±2\sqrt{15}i}{2} when ± is minus.

$$b=-\sqrt{15}i$$

The equation is now solved.

$$b=\sqrt{15}i b=-\sqrt{15}i$$