Solve 9x^2-x-3

$$9 x ^ { 2 } - x - 3$$

Factor

$$9\left(x-\frac{1-\sqrt{109}}{18}\right)\left(x-\frac{\sqrt{109}+1}{18}\right)$$

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.

$$9x^{2}-x-3=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 9\left(-3\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-1\right)±\sqrt{1-36\left(-3\right)}}{2\times 9}$$

Multiply -36 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+108}}{2\times 9}$$

Add 1 to 108.

$$x=\frac{-\left(-1\right)±\sqrt{109}}{2\times 9}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{109}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{1±\sqrt{109}}{18}$$

Now solve the equation x=\frac{1±\sqrt{109}}{18} when ± is plus. Add 1 to \sqrt{109}.

$$x=\frac{\sqrt{109}+1}{18}$$

Now solve the equation x=\frac{1±\sqrt{109}}{18} when ± is minus. Subtract \sqrt{109} from 1.

$$x=\frac{1-\sqrt{109}}{18}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{109}}{18} for x_{1} and \frac{1-\sqrt{109}}{18} for x_{2}.

$$9x^{2}-x-3=9\left(x-\frac{\sqrt{109}+1}{18}\right)\left(x-\frac{1-\sqrt{109}}{18}\right)$$

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 9

$$x ^ 2 -\frac{1}{9}x -\frac{1}{3} = 0$$

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

$$r + s = \frac{1}{9} rs = -\frac{1}{3}$$

Two numbers r and s sum up to \frac{1}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{9} = \frac{1}{18}. 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>

$$r = \frac{1}{18} - u s = \frac{1}{18} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(\frac{1}{18} - u) (\frac{1}{18} + u) = -\frac{1}{3}$$

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

$$\frac{1}{324} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{1}{324} on both sides

$$-u^2 = -\frac{1}{3}-\frac{1}{324} = -\frac{109}{324}$$

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

$$u^2 = \frac{109}{324} u = \pm\sqrt{\frac{109}{324}} = \pm \frac{\sqrt{109}}{18}$$

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

$$r =\frac{1}{18} - \frac{\sqrt{109}}{18} = -0.524 s = \frac{1}{18} + \frac{\sqrt{109}}{18} = 0.636$$

Evaluate

$$9x^{2}-x-3$$