Solve a^2-3/2a-sqrt{12}

$$\frac { a ^ { 2 } - 3 } { 2 a - \sqrt { 12 } }$$

Evaluate

$$\frac{a^{2}-3}{2\left(a-\sqrt{3}\right)}$$

Short Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{a^{2}-3}{2a-2\sqrt{3}}$$

Rationalize the denominator of \frac{a^{2}-3}{2a-2\sqrt{3}} by multiplying numerator and denominator by 2a+2\sqrt{3}.

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

Consider \left(2a-2\sqrt{3}\right)\left(2a+2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(2a\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(a^{2}-3\right)\left(2a+2\sqrt{3}\right)}{4a^{2}-\left(-2\sqrt{3}\right)^{2}}$$

Expand \left(-2\sqrt{3}\right)^{2}.

$$\frac{\left(a^{2}-3\right)\left(2a+2\sqrt{3}\right)}{4a^{2}-\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\left(a^{2}-3\right)\left(2a+2\sqrt{3}\right)}{4a^{2}-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{\left(a^{2}-3\right)\left(2a+2\sqrt{3}\right)}{4a^{2}-12}$$

Factor the expressions that are not already factored.

$$\frac{2\left(a+\sqrt{3}\right)\left(a^{2}-3\right)}{4\left(a^{2}-3\right)}$$

Cancel out 2\left(a^{2}-3\right) in both numerator and denominator.

$$\frac{a+\sqrt{3}}{2}$$

Differentiate w.r.t. a

$$\frac{1}{2} = 0.5$$