Solve 8x^2+16x-3184

$$8 x ^ { 2 } + 16 x - 3184$$

Factor

$$8\left(x-\left(-\sqrt{399}-1\right)\right)\left(x-\left(\sqrt{399}-1\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$8x^{2}+16x-3184=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\left(-3184\right)}}{2\times 8}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 8\left(-3184\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-16±\sqrt{256-32\left(-3184\right)}}{2\times 8}$$

Multiply -32 times -3184.

$$x=\frac{-16±\sqrt{256+101888}}{2\times 8}$$

Add 256 to 101888.

$$x=\frac{-16±\sqrt{102144}}{2\times 8}$$

Take the square root of 102144.

$$x=\frac{-16±16\sqrt{399}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-16±16\sqrt{399}}{16}$$

Now solve the equation x=\frac{-16±16\sqrt{399}}{16} when ± is plus. Add -16 to 16\sqrt{399}.

$$x=\frac{16\sqrt{399}-16}{16}$$

Divide -16+16\sqrt{399} by 16.

$$x=\sqrt{399}-1$$

Now solve the equation x=\frac{-16±16\sqrt{399}}{16} when ± is minus. Subtract 16\sqrt{399} from -16.

$$x=\frac{-16\sqrt{399}-16}{16}$$

Divide -16-16\sqrt{399} by 16.

$$x=-\sqrt{399}-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\sqrt{399} for x_{1} and -1-\sqrt{399} for x_{2}.

$$8x^{2}+16x-3184=8\left(x-\left(\sqrt{399}-1\right)\right)\left(x-\left(-\sqrt{399}-1\right)\right)$$

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 8

$$x ^ 2 +2x -398 = 0$$

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

$$r + s = -2 rs = -398$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -398

$$(-1 - u) (-1 + u) = -398$$

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

$$1 - u^2 = -398$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -398-1 = -399$$

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

$$u^2 = 399 u = \pm\sqrt{399} = \pm \sqrt{399}$$

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

$$r =-1 - \sqrt{399} = -20.975 s = -1 + \sqrt{399} = 18.975$$

Evaluate

$$8\left(x^{2}+2x-398\right)$$