Solve 6x^2-13x+39

$$6 x ^ { 2 } - 13 x + 39$$

Evaluate

$$6x^{2}-13x+39$$

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 6

$$x ^ 2 -\frac{13}{6}x +\frac{13}{2} = 0$$

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

$$r + s = \frac{13}{6} rs = \frac{13}{2}$$

Two numbers r and s sum up to \frac{13}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{6} = \frac{13}{12}. 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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{13}{12} - u s = \frac{13}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{13}{2}

$$(\frac{13}{12} - u) (\frac{13}{12} + u) = \frac{13}{2}$$

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

$$\frac{169}{144} - u^2 = \frac{13}{2}$$

Simplify the expression by subtracting \frac{169}{144} on both sides

$$-u^2 = \frac{13}{2}-\frac{169}{144} = \frac{767}{144}$$

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

$$u^2 = -\frac{767}{144} u = \pm\sqrt{-\frac{767}{144}} = \pm \frac{\sqrt{767}}{12}i$$

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

$$r =\frac{13}{12} - \frac{\sqrt{767}}{12}i = 1.083 - 2.308i s = \frac{13}{12} + \frac{\sqrt{767}}{12}i = 1.083 + 2.308i$$

Differentiate w.r.t. x

$$12x-13$$

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}.

$$2\times 6x^{2-1}-13x^{1-1}$$

Multiply 2 times 6.

$$12x^{2-1}-13x^{1-1}$$

Subtract 1 from 2.

$$12x^{1}-13x^{1-1}$$

Subtract 1 from 1.

$$12x^{1}-13x^{0}$$

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

$$12x-13x^{0}$$

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

$$12x-13$$