Solve 2x^2-7x+3|sqrt{64}

$$2 x ^ { 2 } - 7 x + 3 | \sqrt { 64 }$$

Evaluate

$$2x^{2}-7x+24$$

Solution Steps

Calculate the square root of 64 and get 8.

$$2x^{2}-7x+3|8|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 8 is 8.

$$2x^{2}-7x+3\times 8$$

Multiply 3 and 8 to get 24.

$$2x^{2}-7x+24$$

Differentiate w.r.t. x

$$4x-7$$

Steps Using Derivative Rule for Sum

Calculate the square root of 64 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-7x+3|8|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 8 is 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-7x+3\times 8)$$

Multiply 3 and 8 to get 24.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-7x+24)$$

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}.

$$2\times 2x^{2-1}-7x^{1-1}$$

Multiply 2 times 2.

$$4x^{2-1}-7x^{1-1}$$

Subtract 1 from 2.

$$4x^{1}-7x^{1-1}$$

Subtract 1 from 1.

$$4x^{1}-7x^{0}$$

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

$$4x-7x^{0}$$

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

$$4x-7$$