Solve 10x^2-4x+2

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

Factor

$$2\left(5x^{2}-2x+1\right)$$

Solution Steps

Factor out 2. Polynomial 5x^{2}-2x+1 is not factored since it does not have any rational roots.

$$2\left(5x^{2}-2x+1\right)$$

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.

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

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.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 10\times 2}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-4\right)±\sqrt{16-40\times 2}}{2\times 10}$$

Multiply -40 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-80}}{2\times 10}$$

Add 16 to -80.

$$x=\frac{-\left(-4\right)±\sqrt{-64}}{2\times 10}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

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

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 10

$$x ^ 2 -\frac{2}{5}x +\frac{1}{5} = 0$$

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

$$r + s = \frac{2}{5} rs = \frac{1}{5}$$

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

$$r = \frac{1}{5} - u s = \frac{1}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}

$$(\frac{1}{5} - u) (\frac{1}{5} + u) = \frac{1}{5}$$

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

$$\frac{1}{25} - u^2 = \frac{1}{5}$$

Simplify the expression by subtracting \frac{1}{25} on both sides

$$-u^2 = \frac{1}{5}-\frac{1}{25} = \frac{4}{25}$$

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

$$u^2 = -\frac{4}{25} u = \pm\sqrt{-\frac{4}{25}} = \pm \frac{2}{5}i$$

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

$$r =\frac{1}{5} - \frac{2}{5}i = 0.200 - 0.400i s = \frac{1}{5} + \frac{2}{5}i = 0.200 + 0.400i$$

Evaluate

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