Solve 5x^2+40x+99

$$5 x ^ { 2 } + 40 x + 99$$

Evaluate

$$5x^{2}+40x+99$$

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 5

$$x ^ 2 +8x +\frac{99}{5} = 0$$

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

$$r + s = -8 rs = \frac{99}{5}$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{99}{5}

$$(-4 - u) (-4 + u) = \frac{99}{5}$$

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

$$16 - u^2 = \frac{99}{5}$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = \frac{99}{5}-16 = \frac{19}{5}$$

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

$$u^2 = -\frac{19}{5} u = \pm\sqrt{-\frac{19}{5}} = \pm \frac{\sqrt{19}}{\sqrt{5}}i$$

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

$$r =-4 - \frac{\sqrt{19}}{\sqrt{5}}i = -4 - 1.949i s = -4 + \frac{\sqrt{19}}{\sqrt{5}}i = -4 + 1.949i$$

Differentiate w.r.t. x

$$10\left(x+4\right)$$

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}.

$$2\times 5x^{2-1}+40x^{1-1}$$

Multiply 2 times 5.

$$10x^{2-1}+40x^{1-1}$$

Subtract 1 from 2.

$$10x^{1}+40x^{1-1}$$

Subtract 1 from 1.

$$10x^{1}+40x^{0}$$

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

$$10x+40x^{0}$$

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

$$10x+40\times 1$$

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

$$10x+40$$