Solve x^2-110x+500

$$x ^ { 2 } - 110 x + 500$$

Factor

$$\left(x-\left(55-5\sqrt{101}\right)\right)\left(x-\left(5\sqrt{101}+55\right)\right)$$

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.

$$x^{2}-110x+500=0$$

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.

$$x=\frac{-\left(-110\right)±\sqrt{\left(-110\right)^{2}-4\times 500}}{2}$$

Square -110.

$$x=\frac{-\left(-110\right)±\sqrt{12100-4\times 500}}{2}$$

Multiply -4 times 500.

$$x=\frac{-\left(-110\right)±\sqrt{12100-2000}}{2}$$

Add 12100 to -2000.

$$x=\frac{-\left(-110\right)±\sqrt{10100}}{2}$$

Take the square root of 10100.

$$x=\frac{-\left(-110\right)±10\sqrt{101}}{2}$$

The opposite of -110 is 110.

$$x=\frac{110±10\sqrt{101}}{2}$$

Now solve the equation x=\frac{110±10\sqrt{101}}{2} when ± is plus. Add 110 to 10\sqrt{101}.

$$x=\frac{10\sqrt{101}+110}{2}$$

Divide 110+10\sqrt{101} by 2.

$$x=5\sqrt{101}+55$$

Now solve the equation x=\frac{110±10\sqrt{101}}{2} when ± is minus. Subtract 10\sqrt{101} from 110.

$$x=\frac{110-10\sqrt{101}}{2}$$

Divide 110-10\sqrt{101} by 2.

$$x=55-5\sqrt{101}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 55+5\sqrt{101} for x_{1} and 55-5\sqrt{101} for x_{2}.

$$x^{2}-110x+500=\left(x-\left(5\sqrt{101}+55\right)\right)\left(x-\left(55-5\sqrt{101}\right)\right)$$

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.

$$x ^ 2 -110x +500 = 0$$

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

$$r + s = 110 rs = 500$$

Two numbers r and s sum up to 110 exactly when the average of the two numbers is \frac{1}{2}*110 = 55. 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>

$$r = 55 - u s = 55 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 500

$$(55 - u) (55 + u) = 500$$

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

$$3025 - u^2 = 500$$

Simplify the expression by subtracting 3025 on both sides

$$-u^2 = 500-3025 = -2525$$

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

$$u^2 = 2525 u = \pm\sqrt{2525} = \pm \sqrt{2525}$$

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

$$r =55 - \sqrt{2525} = 4.751 s = 55 + \sqrt{2525} = 105.249$$

Evaluate

$$x^{2}-110x+500$$