Solve -x^2+20x-1

$$- x ^ { 2 } + 20 x - 1$$

Factor

$$-\left(x-\left(10-3\sqrt{11}\right)\right)\left(x-\left(3\sqrt{11}+10\right)\right)$$

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.

$$-x^{2}+20x-1=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-20±\sqrt{400+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

$$x=\frac{-20±\sqrt{400-4}}{2\left(-1\right)}$$

Add 400 to -4.

$$x=\frac{-20±\sqrt{396}}{2\left(-1\right)}$$

Take the square root of 396.

$$x=\frac{-20±6\sqrt{11}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-20±6\sqrt{11}}{-2}$$

Now solve the equation x=\frac{-20±6\sqrt{11}}{-2} when ± is plus. Add -20 to 6\sqrt{11}.

$$x=\frac{6\sqrt{11}-20}{-2}$$

Divide -20+6\sqrt{11} by -2.

$$x=10-3\sqrt{11}$$

Now solve the equation x=\frac{-20±6\sqrt{11}}{-2} when ± is minus. Subtract 6\sqrt{11} from -20.

$$x=\frac{-6\sqrt{11}-20}{-2}$$

Divide -20-6\sqrt{11} by -2.

$$x=3\sqrt{11}+10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10-3\sqrt{11} for x_{1} and 10+3\sqrt{11} for x_{2}.

$$-x^{2}+20x-1=-\left(x-\left(10-3\sqrt{11}\right)\right)\left(x-\left(3\sqrt{11}+10\right)\right)$$

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.

$$x ^ 2 -20x +1 = 0$$

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

$$r + s = 20 rs = 1$$

Two numbers r and s sum up to 20 exactly when the average of the two numbers is \frac{1}{2}*20 = 10. 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>

$$r = 10 - u s = 10 + u$$

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

$$(10 - u) (10 + u) = 1$$

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

$$100 - u^2 = 1$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = 1-100 = -99$$

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

$$u^2 = 99 u = \pm\sqrt{99} = \pm \sqrt{99}$$

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

$$r =10 - \sqrt{99} = 0.050 s = 10 + \sqrt{99} = 19.950$$

Evaluate

$$-x^{2}+20x-1$$