Solve 225m^2+30m+9

$$225 m ^ { 2 } + 30 m + 9$$

Factor

$$3\left(75m^{2}+10m+3\right)$$

Solution Steps

Factor out 3. Polynomial 75m^{2}+10m+3 is not factored since it does not have any rational roots.

$$3\left(75m^{2}+10m+3\right)$$

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.

$$225m^{2}+30m+9=0$$

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.

$$m=\frac{-30±\sqrt{30^{2}-4\times 225\times 9}}{2\times 225}$$

Square 30.

$$m=\frac{-30±\sqrt{900-4\times 225\times 9}}{2\times 225}$$

Multiply -4 times 225.

$$m=\frac{-30±\sqrt{900-900\times 9}}{2\times 225}$$

Multiply -900 times 9.

$$m=\frac{-30±\sqrt{900-8100}}{2\times 225}$$

Add 900 to -8100.

$$m=\frac{-30±\sqrt{-7200}}{2\times 225}$$

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

$$225m^{2}+30m+9$$

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 225

$$x ^ 2 +\frac{2}{15}x +\frac{1}{25} = 0$$

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

$$r + s = -\frac{2}{15} rs = \frac{1}{25}$$

Two numbers r and s sum up to -\frac{2}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{15} = -\frac{1}{15}. 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>

$$r = -\frac{1}{15} - u s = -\frac{1}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{25}

$$(-\frac{1}{15} - u) (-\frac{1}{15} + u) = \frac{1}{25}$$

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

$$\frac{1}{225} - u^2 = \frac{1}{25}$$

Simplify the expression by subtracting \frac{1}{225} on both sides

$$-u^2 = \frac{1}{25}-\frac{1}{225} = \frac{8}{225}$$

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

$$u^2 = -\frac{8}{225} u = \pm\sqrt{-\frac{8}{225}} = \pm \frac{\sqrt{8}}{15}i$$

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

$$r =-\frac{1}{15} - \frac{\sqrt{8}}{15}i = -0.067 - 0.189i s = -\frac{1}{15} + \frac{\sqrt{8}}{15}i = -0.067 + 0.189i$$

Evaluate

$$225m^{2}+30m+9$$