Solve (2x+sqrt{5})(x+sqrt{5})

$$( 2 x + \sqrt { 5 } ) ( x + \sqrt { 5 } )$$

Evaluate

$$\left(x+\sqrt{5}\right)\left(2x+\sqrt{5}\right)$$

Solution Steps

Apply the distributive property by multiplying each term of 2x+\sqrt{5} by each term of x+\sqrt{5}.

$$2x^{2}+2x\sqrt{5}+\sqrt{5}x+\left(\sqrt{5}\right)^{2}$$

Combine 2x\sqrt{5} and \sqrt{5}x to get 3x\sqrt{5}.

$$2x^{2}+3x\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$2x^{2}+3x\sqrt{5}+5$$

Differentiate w.r.t. x

$$4x+3\sqrt{5}$$

Steps Using Derivative Rule for Sum

Apply the distributive property by multiplying each term of 2x+\sqrt{5} by each term of x+\sqrt{5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+2x\sqrt{5}+\sqrt{5}x+\left(\sqrt{5}\right)^{2})$$

Combine 2x\sqrt{5} and \sqrt{5}x to get 3x\sqrt{5}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+3x\sqrt{5}+\left(\sqrt{5}\right)^{2})$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+3x\sqrt{5}+5)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 2x^{2-1}+3\sqrt{5}x^{1-1}$$

Multiply 2 times 2.

$$4x^{2-1}+3\sqrt{5}x^{1-1}$$

Subtract 1 from 2.

$$4x^{1}+3\sqrt{5}x^{1-1}$$

Subtract 1 from 1.

$$4x^{1}+3\sqrt{5}x^{0}$$

For any term t, t^{1}=t.

$$4x+3\sqrt{5}x^{0}$$

For any term t except 0, t^{0}=1.

$$4x+3\sqrt{5}\times 1$$

For any term t, t\times 1=t and 1t=t.

$$4x+3\sqrt{5}$$