Solve -x^2+12x-31

$$- x ^ { 2 } + 12 x - 31$$

Factor

$$-\left(x-\left(6-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+6\right)\right)$$

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.

$$-x^{2}+12x-31=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-1\right)\left(-31\right)}}{2\left(-1\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-1\right)\left(-31\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-12±\sqrt{144+4\left(-31\right)}}{2\left(-1\right)}$$

Multiply 4 times -31.

$$x=\frac{-12±\sqrt{144-124}}{2\left(-1\right)}$$

Add 144 to -124.

$$x=\frac{-12±\sqrt{20}}{2\left(-1\right)}$$

Take the square root of 20.

$$x=\frac{-12±2\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-12±2\sqrt{5}}{-2}$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{-2} when ± is plus. Add -12 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-12}{-2}$$

Divide -12+2\sqrt{5} by -2.

$$x=6-\sqrt{5}$$

Now solve the equation x=\frac{-12±2\sqrt{5}}{-2} when ± is minus. Subtract 2\sqrt{5} from -12.

$$x=\frac{-2\sqrt{5}-12}{-2}$$

Divide -12-2\sqrt{5} by -2.

$$x=\sqrt{5}+6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6-\sqrt{5} for x_{1} and 6+\sqrt{5} for x_{2}.

$$-x^{2}+12x-31=-\left(x-\left(6-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+6\right)\right)$$

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.

$$x ^ 2 -12x +31 = 0$$

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

$$r + s = 12 rs = 31$$

Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 6 - u s = 6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 31

$$(6 - u) (6 + u) = 31$$

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

$$36 - u^2 = 31$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 31-36 = -5$$

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

$$u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{5}$$

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

$$r =6 - \sqrt{5} = 3.764 s = 6 + \sqrt{5} = 8.236$$

Evaluate

$$-x^{2}+12x-31$$