Solve 9x^2-6x+15

$$9 x ^ { 2 } - 6 x + 15$$

Factor

$$3\left(3x^{2}-2x+5\right)$$

Solution Steps

Factor out 3. Polynomial 3x^{2}-2x+5 is not factored since it does not have any rational roots.

$$3\left(3x^{2}-2x+5\right)$$

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.

$$9x^{2}-6x+15=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\times 15}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\times 15}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\times 15}}{2\times 9}$$

Multiply -36 times 15.

$$x=\frac{-\left(-6\right)±\sqrt{36-540}}{2\times 9}$$

Add 36 to -540.

$$x=\frac{-\left(-6\right)±\sqrt{-504}}{2\times 9}$$

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

$$9x^{2}-6x+15$$

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 9

$$x ^ 2 -\frac{2}{3}x +\frac{5}{3} = 0$$

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

$$r + s = \frac{2}{3} rs = \frac{5}{3}$$

Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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>

$$r = \frac{1}{3} - u s = \frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = \frac{5}{3}$$

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

$$\frac{1}{9} - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = \frac{5}{3}-\frac{1}{9} = \frac{14}{9}$$

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

$$u^2 = -\frac{14}{9} u = \pm\sqrt{-\frac{14}{9}} = \pm \frac{\sqrt{14}}{3}i$$

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

$$r =\frac{1}{3} - \frac{\sqrt{14}}{3}i = 0.333 - 1.247i s = \frac{1}{3} + \frac{\sqrt{14}}{3}i = 0.333 + 1.247i$$

Evaluate

$$9x^{2}-6x+15$$