Solve 8x^2-2x-8

$$8 x ^ { 2 } - 2 x - 8$$

Factor

$$8\left(x-\frac{1-\sqrt{65}}{8}\right)\left(x-\frac{\sqrt{65}+1}{8}\right)$$

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.

$$8x^{2}-2x-8=0$$

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.

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

Square -2.

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

Multiply -4 times 8.

$$x=\frac{-\left(-2\right)±\sqrt{4-32\left(-8\right)}}{2\times 8}$$

Multiply -32 times -8.

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

Add 4 to 256.

$$x=\frac{-\left(-2\right)±\sqrt{260}}{2\times 8}$$

Take the square root of 260.

$$x=\frac{-\left(-2\right)±2\sqrt{65}}{2\times 8}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{65}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{2±2\sqrt{65}}{16}$$

Now solve the equation x=\frac{2±2\sqrt{65}}{16} when ± is plus. Add 2 to 2\sqrt{65}.

$$x=\frac{2\sqrt{65}+2}{16}$$

Divide 2+2\sqrt{65} by 16.

$$x=\frac{\sqrt{65}+1}{8}$$

Now solve the equation x=\frac{2±2\sqrt{65}}{16} when ± is minus. Subtract 2\sqrt{65} from 2.

$$x=\frac{2-2\sqrt{65}}{16}$$

Divide 2-2\sqrt{65} by 16.

$$x=\frac{1-\sqrt{65}}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{65}}{8} for x_{1} and \frac{1-\sqrt{65}}{8} for x_{2}.

$$8x^{2}-2x-8=8\left(x-\frac{\sqrt{65}+1}{8}\right)\left(x-\frac{1-\sqrt{65}}{8}\right)$$

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

$$x ^ 2 -\frac{1}{4}x -1 = 0$$

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

$$r + s = \frac{1}{4} rs = -1$$

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

$$r = \frac{1}{8} - u s = \frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(\frac{1}{8} - u) (\frac{1}{8} + u) = -1$$

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

$$\frac{1}{64} - u^2 = -1$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -1-\frac{1}{64} = -\frac{65}{64}$$

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

$$u^2 = \frac{65}{64} u = \pm\sqrt{\frac{65}{64}} = \pm \frac{\sqrt{65}}{8}$$

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

$$r =\frac{1}{8} - \frac{\sqrt{65}}{8} = -0.883 s = \frac{1}{8} + \frac{\sqrt{65}}{8} = 1.133$$

Evaluate

$$8x^{2}-2x-8$$