Solve 18x^2-6x+48

$$18 x ^ { 2 } - 6 x + 48$$

Factor

$$6\left(3x^{2}-x+8\right)$$

Solution Steps

Factor out 6. Polynomial 3x^{2}-x+8 is not factored since it does not have any rational roots.

$$6\left(3x^{2}-x+8\right)$$

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.

$$18x^{2}-6x+48=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 18\times 48}}{2\times 18}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 18\times 48}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-6\right)±\sqrt{36-72\times 48}}{2\times 18}$$

Multiply -72 times 48.

$$x=\frac{-\left(-6\right)±\sqrt{36-3456}}{2\times 18}$$

Add 36 to -3456.

$$x=\frac{-\left(-6\right)±\sqrt{-3420}}{2\times 18}$$

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

$$18x^{2}-6x+48$$

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 18

$$x ^ 2 -\frac{1}{3}x +\frac{8}{3} = 0$$

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

$$r + s = \frac{1}{3} rs = \frac{8}{3}$$

Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{1}{6} - u s = \frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{3}

$$(\frac{1}{6} - u) (\frac{1}{6} + u) = \frac{8}{3}$$

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

$$\frac{1}{36} - u^2 = \frac{8}{3}$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = \frac{8}{3}-\frac{1}{36} = \frac{95}{36}$$

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

$$u^2 = -\frac{95}{36} u = \pm\sqrt{-\frac{95}{36}} = \pm \frac{\sqrt{95}}{6}i$$

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

$$r =\frac{1}{6} - \frac{\sqrt{95}}{6}i = 0.167 - 1.624i s = \frac{1}{6} + \frac{\sqrt{95}}{6}i = 0.167 + 1.624i$$

Evaluate

$$18x^{2}-6x+48$$