Solve f(x)=-9x^2+3x+8

$$f ( x ) = - 9 x ^ { 2 } + 3 x + 8$$

Factor

$$-9\left(x-\frac{1-\sqrt{33}}{6}\right)\left(x-\frac{\sqrt{33}+1}{6}\right)$$

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.

$$-9x^{2}+3x+8=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-9\right)\times 8}}{2\left(-9\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-9\right)\times 8}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-3±\sqrt{9+36\times 8}}{2\left(-9\right)}$$

Multiply 36 times 8.

$$x=\frac{-3±\sqrt{9+288}}{2\left(-9\right)}$$

Add 9 to 288.

$$x=\frac{-3±\sqrt{297}}{2\left(-9\right)}$$

Take the square root of 297.

$$x=\frac{-3±3\sqrt{33}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-3±3\sqrt{33}}{-18}$$

Now solve the equation x=\frac{-3±3\sqrt{33}}{-18} when ± is plus. Add -3 to 3\sqrt{33}.

$$x=\frac{3\sqrt{33}-3}{-18}$$

Divide -3+3\sqrt{33} by -18.

$$x=\frac{1-\sqrt{33}}{6}$$

Now solve the equation x=\frac{-3±3\sqrt{33}}{-18} when ± is minus. Subtract 3\sqrt{33} from -3.

$$x=\frac{-3\sqrt{33}-3}{-18}$$

Divide -3-3\sqrt{33} by -18.

$$x=\frac{\sqrt{33}+1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1-\sqrt{33}}{6} for x_{1} and \frac{1+\sqrt{33}}{6} for x_{2}.

$$-9x^{2}+3x+8=-9\left(x-\frac{1-\sqrt{33}}{6}\right)\left(x-\frac{\sqrt{33}+1}{6}\right)$$

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.

$$x ^ 2 -\frac{1}{3}x -\frac{8}{9} = 0$$

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

$$r + s = \frac{1}{3} rs = -\frac{8}{9}$$

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}{9}

$$(\frac{1}{6} - u) (\frac{1}{6} + u) = -\frac{8}{9}$$

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

$$\frac{1}{36} - u^2 = -\frac{8}{9}$$

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

$$-u^2 = -\frac{8}{9}-\frac{1}{36} = -\frac{11}{12}$$

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

$$u^2 = \frac{11}{12} u = \pm\sqrt{\frac{11}{12}} = \pm \frac{\sqrt{11}}{\sqrt{12}}$$

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

$$r =\frac{1}{6} - \frac{\sqrt{11}}{\sqrt{12}} = -0.791 s = \frac{1}{6} + \frac{\sqrt{11}}{\sqrt{12}} = 1.124$$

Evaluate

$$8+3x-9x^{2}$$