Solve (i)(sqrt{3}-sqrt{2})^2

$$( i ) ( \sqrt { 3 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$-2\sqrt{6}i+5i\approx 0.101020514i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.

$$i\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$i\left(3-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$i\left(3-2\sqrt{6}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$i\left(3-2\sqrt{6}+2\right)$$

Add 3 and 2 to get 5.

$$i\left(5-2\sqrt{6}\right)$$

Use the distributive property to multiply i by 5-2\sqrt{6}.

$$5i-2i\sqrt{6}$$

Expand

$$-2\sqrt{6}i+5i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.

$$i\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$i\left(3-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$i\left(3-2\sqrt{6}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$i\left(3-2\sqrt{6}+2\right)$$

Add 3 and 2 to get 5.

$$i\left(5-2\sqrt{6}\right)$$

Use the distributive property to multiply i by 5-2\sqrt{6}.

$$5i-2i\sqrt{6}$$