Solve 65x^2-3x+4

$$65 x ^ { 2 } - 3 x + 4$$

Evaluate

$$65x^{2}-3x+4$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 65

$$x ^ 2 -\frac{3}{65}x +\frac{4}{65} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{3}{65} rs = \frac{4}{65}$$

Two numbers r and s sum up to \frac{3}{65} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{65} = \frac{3}{130}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{3}{130} - u s = \frac{3}{130} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{65}

$$(\frac{3}{130} - u) (\frac{3}{130} + u) = \frac{4}{65}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{16900} - u^2 = \frac{4}{65}$$

Simplify the expression by subtracting \frac{9}{16900} on both sides

$$-u^2 = \frac{4}{65}-\frac{9}{16900} = \frac{1031}{16900}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = -\frac{1031}{16900} u = \pm\sqrt{-\frac{1031}{16900}} = \pm \frac{\sqrt{1031}}{130}i$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{3}{130} - \frac{\sqrt{1031}}{130}i = 0.023 - 0.247i s = \frac{3}{130} + \frac{\sqrt{1031}}{130}i = 0.023 + 0.247i$$

Differentiate w.r.t. x

$$130x-3$$

Steps Using Derivative Rule for Sum

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}.

$$2\times 65x^{2-1}-3x^{1-1}$$

Multiply 2 times 65.

$$130x^{2-1}-3x^{1-1}$$

Subtract 1 from 2.

$$130x^{1}-3x^{1-1}$$

Subtract 1 from 1.

$$130x^{1}-3x^{0}$$

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

$$130x-3x^{0}$$

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

$$130x-3$$