Solve x^2+64x+8

$$x ^ { 2 } + 64 x + 8$$

Factor

$$\left(x-\left(-2\sqrt{254}-32\right)\right)\left(x-\left(2\sqrt{254}-32\right)\right)$$

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.

$$x^{2}+64x+8=0$$

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.

$$x=\frac{-64±\sqrt{64^{2}-4\times 8}}{2}$$

Square 64.

$$x=\frac{-64±\sqrt{4096-4\times 8}}{2}$$

Multiply -4 times 8.

$$x=\frac{-64±\sqrt{4096-32}}{2}$$

Add 4096 to -32.

$$x=\frac{-64±\sqrt{4064}}{2}$$

Take the square root of 4064.

$$x=\frac{-64±4\sqrt{254}}{2}$$

Now solve the equation x=\frac{-64±4\sqrt{254}}{2} when ± is plus. Add -64 to 4\sqrt{254}.

$$x=\frac{4\sqrt{254}-64}{2}$$

Divide -64+4\sqrt{254} by 2.

$$x=2\sqrt{254}-32$$

Now solve the equation x=\frac{-64±4\sqrt{254}}{2} when ± is minus. Subtract 4\sqrt{254} from -64.

$$x=\frac{-4\sqrt{254}-64}{2}$$

Divide -64-4\sqrt{254} by 2.

$$x=-2\sqrt{254}-32$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -32+2\sqrt{254} for x_{1} and -32-2\sqrt{254} for x_{2}.

$$x^{2}+64x+8=\left(x-\left(2\sqrt{254}-32\right)\right)\left(x-\left(-2\sqrt{254}-32\right)\right)$$

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.

$$x ^ 2 +64x +8 = 0$$

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

$$r + s = -64 rs = 8$$

Two numbers r and s sum up to -64 exactly when the average of the two numbers is \frac{1}{2}*-64 = -32. 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>

$$r = -32 - u s = -32 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 8

$$(-32 - u) (-32 + u) = 8$$

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

$$1024 - u^2 = 8$$

Simplify the expression by subtracting 1024 on both sides

$$-u^2 = 8-1024 = -1016$$

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

$$u^2 = 1016 u = \pm\sqrt{1016} = \pm \sqrt{1016}$$

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

$$r =-32 - \sqrt{1016} = -63.875 s = -32 + \sqrt{1016} = -0.125$$

Evaluate

$$x^{2}+64x+8$$