Solve x^2+x+210

$$x ^ { 2 } + x + 210$$

Evaluate

$$x^{2}+x+210$$

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.

$$x ^ 2 +1x +210 = 0$$

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

$$r + s = -1 rs = 210$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 210

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = 210$$

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

$$\frac{1}{4} - u^2 = 210$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = 210-\frac{1}{4} = \frac{839}{4}$$

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

$$u^2 = -\frac{839}{4} u = \pm\sqrt{-\frac{839}{4}} = \pm \frac{\sqrt{839}}{2}i$$

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

$$r =-\frac{1}{2} - \frac{\sqrt{839}}{2}i = -0.500 - 14.483i s = -\frac{1}{2} + \frac{\sqrt{839}}{2}i = -0.500 + 14.483i$$

Differentiate w.r.t. x

$$2x+1$$

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}.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

$$2x^{1}+x^{0}$$

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

$$2x+x^{0}$$

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

$$2x+1$$