Solve -7x^2+5x-4

$$- 7 x ^ { 2 } + 5 x - 4$$

Evaluate

$$-7x^{2}+5x-4$$

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.

$$x ^ 2 -\frac{5}{7}x +\frac{4}{7} = 0$$

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

$$r + s = \frac{5}{7} rs = \frac{4}{7}$$

Two numbers r and s sum up to \frac{5}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{7} = \frac{5}{14}. 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>

$$r = \frac{5}{14} - u s = \frac{5}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{7}

$$(\frac{5}{14} - u) (\frac{5}{14} + u) = \frac{4}{7}$$

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

$$\frac{25}{196} - u^2 = \frac{4}{7}$$

Simplify the expression by subtracting \frac{25}{196} on both sides

$$-u^2 = \frac{4}{7}-\frac{25}{196} = \frac{87}{196}$$

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

$$u^2 = -\frac{87}{196} u = \pm\sqrt{-\frac{87}{196}} = \pm \frac{\sqrt{87}}{14}i$$

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

$$r =\frac{5}{14} - \frac{\sqrt{87}}{14}i = 0.357 - 0.666i s = \frac{5}{14} + \frac{\sqrt{87}}{14}i = 0.357 + 0.666i$$

Differentiate w.r.t. x

$$5-14x$$

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}.

$$2\left(-7\right)x^{2-1}+5x^{1-1}$$

Multiply 2 times -7.

$$-14x^{2-1}+5x^{1-1}$$

Subtract 1 from 2.

$$-14x^{1}+5x^{1-1}$$

Subtract 1 from 1.

$$-14x^{1}+5x^{0}$$

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

$$-14x+5x^{0}$$

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

$$-14x+5\times 1$$

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

$$-14x+5$$