Solve 9x^2+8x-9

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

Factor

$$9\left(x-\frac{-\sqrt{97}-4}{9}\right)\left(x-\frac{\sqrt{97}-4}{9}\right)$$

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.

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

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.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 9\left(-9\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-8±\sqrt{64-36\left(-9\right)}}{2\times 9}$$

Multiply -36 times -9.

$$x=\frac{-8±\sqrt{64+324}}{2\times 9}$$

Add 64 to 324.

$$x=\frac{-8±\sqrt{388}}{2\times 9}$$

Take the square root of 388.

$$x=\frac{-8±2\sqrt{97}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-8±2\sqrt{97}}{18}$$

Now solve the equation x=\frac{-8±2\sqrt{97}}{18} when ± is plus. Add -8 to 2\sqrt{97}.

$$x=\frac{2\sqrt{97}-8}{18}$$

Divide -8+2\sqrt{97} by 18.

$$x=\frac{\sqrt{97}-4}{9}$$

Now solve the equation x=\frac{-8±2\sqrt{97}}{18} when ± is minus. Subtract 2\sqrt{97} from -8.

$$x=\frac{-2\sqrt{97}-8}{18}$$

Divide -8-2\sqrt{97} by 18.

$$x=\frac{-\sqrt{97}-4}{9}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{97}}{9} for x_{1} and \frac{-4-\sqrt{97}}{9} for x_{2}.

$$9x^{2}+8x-9=9\left(x-\frac{\sqrt{97}-4}{9}\right)\left(x-\frac{-\sqrt{97}-4}{9}\right)$$

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

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

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

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

Two numbers r and s sum up to -\frac{8}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{9} = -\frac{4}{9}. 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>

$$r = -\frac{4}{9} - u s = -\frac{4}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(-\frac{4}{9} - u) (-\frac{4}{9} + u) = -1$$

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

$$\frac{16}{81} - u^2 = -1$$

Simplify the expression by subtracting \frac{16}{81} on both sides

$$-u^2 = -1-\frac{16}{81} = -\frac{97}{81}$$

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

$$u^2 = \frac{97}{81} u = \pm\sqrt{\frac{97}{81}} = \pm \frac{\sqrt{97}}{9}$$

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

$$r =-\frac{4}{9} - \frac{\sqrt{97}}{9} = -1.539 s = -\frac{4}{9} + \frac{\sqrt{97}}{9} = 0.650$$

Evaluate

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