Solve 9x^2-15x-0

$$9 x ^ { 2 } - 15 x - 0$$

Evaluate

$$3x\left(3x-5\right)$$

Solution Steps

Multiply -1 and 0 to get 0.

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

Anything plus zero gives itself.

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

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

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

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

$$r + s = \frac{5}{3} rs = 0$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{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>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 0

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = 0$$

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

$$\frac{25}{36} - u^2 = 0$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = 0-\frac{25}{36} = -\frac{25}{36}$$

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

$$u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}$$

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

$$r =\frac{5}{6} - \frac{5}{6} = 0 s = \frac{5}{6} + \frac{5}{6} = 1.667$$

Factor

$$3x\left(3x-5\right)$$

Solution Steps

Factor out 3.

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

Consider 3x^{2}-5x. Factor out x.

$$x\left(3x-5\right)$$

Rewrite the complete factored expression.

$$3x\left(3x-5\right)$$

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.

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

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.

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

Take the square root of \left(-15\right)^{2}.

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

The opposite of -15 is 15.

$$x=\frac{15±15}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{15±15}{18}$$

Now solve the equation x=\frac{15±15}{18} when ± is plus. Add 15 to 15.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{15±15}{18} when ± is minus. Subtract 15 from 15.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{3} for x_{1} and 0 for x_{2}.

$$9x^{2}-15x=9\left(x-\frac{5}{3}\right)x$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-15x=9\times \frac{3x-5}{3}x$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9x^{2}-15x=3\left(3x-5\right)x$$