Solve a^2-15a-36

$$a ^ { 2 } - 15 a - 36$$

Factor

$$\left(a-\frac{15-3\sqrt{41}}{2}\right)\left(a-\frac{3\sqrt{41}+15}{2}\right)$$

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.

$$a^{2}-15a-36=0$$

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.

$$a=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-36\right)}}{2}$$

Square -15.

$$a=\frac{-\left(-15\right)±\sqrt{225-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$a=\frac{-\left(-15\right)±\sqrt{225+144}}{2}$$

Add 225 to 144.

$$a=\frac{-\left(-15\right)±\sqrt{369}}{2}$$

Take the square root of 369.

$$a=\frac{-\left(-15\right)±3\sqrt{41}}{2}$$

The opposite of -15 is 15.

$$a=\frac{15±3\sqrt{41}}{2}$$

Now solve the equation a=\frac{15±3\sqrt{41}}{2} when ± is plus. Add 15 to 3\sqrt{41}.

$$a=\frac{3\sqrt{41}+15}{2}$$

Now solve the equation a=\frac{15±3\sqrt{41}}{2} when ± is minus. Subtract 3\sqrt{41} from 15.

$$a=\frac{15-3\sqrt{41}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{15+3\sqrt{41}}{2} for x_{1} and \frac{15-3\sqrt{41}}{2} for x_{2}.

$$a^{2}-15a-36=\left(a-\frac{3\sqrt{41}+15}{2}\right)\left(a-\frac{15-3\sqrt{41}}{2}\right)$$

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.

$$x ^ 2 -15x -36 = 0$$

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

$$r + s = 15 rs = -36$$

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

$$r = \frac{15}{2} - u s = \frac{15}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -36

$$(\frac{15}{2} - u) (\frac{15}{2} + u) = -36$$

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

$$\frac{225}{4} - u^2 = -36$$

Simplify the expression by subtracting \frac{225}{4} on both sides

$$-u^2 = -36-\frac{225}{4} = -\frac{369}{4}$$

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

$$u^2 = \frac{369}{4} u = \pm\sqrt{\frac{369}{4}} = \pm \frac{\sqrt{369}}{2}$$

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

$$r =\frac{15}{2} - \frac{\sqrt{369}}{2} = -2.105 s = \frac{15}{2} + \frac{\sqrt{369}}{2} = 17.105$$

Evaluate

$$a^{2}-15a-36$$