Solve 1006x^2-10x-4

$$1006 x ^ { 2 } - 10 x - 4$$

Factor

$$1006\left(x-\frac{5-\sqrt{4049}}{1006}\right)\left(x-\frac{\sqrt{4049}+5}{1006}\right)$$

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.

$$1006x^{2}-10x-4=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1006\left(-4\right)}}{2\times 1006}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 1006\left(-4\right)}}{2\times 1006}$$

Multiply -4 times 1006.

$$x=\frac{-\left(-10\right)±\sqrt{100-4024\left(-4\right)}}{2\times 1006}$$

Multiply -4024 times -4.

$$x=\frac{-\left(-10\right)±\sqrt{100+16096}}{2\times 1006}$$

Add 100 to 16096.

$$x=\frac{-\left(-10\right)±\sqrt{16196}}{2\times 1006}$$

Take the square root of 16196.

$$x=\frac{-\left(-10\right)±2\sqrt{4049}}{2\times 1006}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{4049}}{2\times 1006}$$

Multiply 2 times 1006.

$$x=\frac{10±2\sqrt{4049}}{2012}$$

Now solve the equation x=\frac{10±2\sqrt{4049}}{2012} when ± is plus. Add 10 to 2\sqrt{4049}.

$$x=\frac{2\sqrt{4049}+10}{2012}$$

Divide 10+2\sqrt{4049} by 2012.

$$x=\frac{\sqrt{4049}+5}{1006}$$

Now solve the equation x=\frac{10±2\sqrt{4049}}{2012} when ± is minus. Subtract 2\sqrt{4049} from 10.

$$x=\frac{10-2\sqrt{4049}}{2012}$$

Divide 10-2\sqrt{4049} by 2012.

$$x=\frac{5-\sqrt{4049}}{1006}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{4049}}{1006} for x_{1} and \frac{5-\sqrt{4049}}{1006} for x_{2}.

$$1006x^{2}-10x-4=1006\left(x-\frac{\sqrt{4049}+5}{1006}\right)\left(x-\frac{5-\sqrt{4049}}{1006}\right)$$

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 1006

$$x ^ 2 -\frac{5}{503}x -\frac{2}{503} = 0$$

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

$$r + s = \frac{5}{503} rs = -\frac{2}{503}$$

Two numbers r and s sum up to \frac{5}{503} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{503} = \frac{5}{1006}. 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>

$$r = \frac{5}{1006} - u s = \frac{5}{1006} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{503}

$$(\frac{5}{1006} - u) (\frac{5}{1006} + u) = -\frac{2}{503}$$

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

$$\frac{25}{1012036} - u^2 = -\frac{2}{503}$$

Simplify the expression by subtracting \frac{25}{1012036} on both sides

$$-u^2 = -\frac{2}{503}-\frac{25}{1012036} = \frac{4049}{1012036}$$

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

$$u^2 = -\frac{4049}{1012036} u = \pm\sqrt{-\frac{4049}{1012036}} = \pm \frac{\sqrt{4049}}{1006}i$$

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

$$r =\frac{5}{1006} - \frac{\sqrt{4049}}{1006}i = -0.058 s = \frac{5}{1006} + \frac{\sqrt{4049}}{1006}i = 0.068$$

Evaluate

$$1006x^{2}-10x-4$$