Solve x^2-60x+540

$$x ^ { 2 } - 60 x + 540$$

Factor

$$\left(x-\left(30-6\sqrt{10}\right)\right)\left(x-\left(6\sqrt{10}+30\right)\right)$$

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.

$$x^{2}-60x+540=0$$

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.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 540}}{2}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 540}}{2}$$

Multiply -4 times 540.

$$x=\frac{-\left(-60\right)±\sqrt{3600-2160}}{2}$$

Add 3600 to -2160.

$$x=\frac{-\left(-60\right)±\sqrt{1440}}{2}$$

Take the square root of 1440.

$$x=\frac{-\left(-60\right)±12\sqrt{10}}{2}$$

The opposite of -60 is 60.

$$x=\frac{60±12\sqrt{10}}{2}$$

Now solve the equation x=\frac{60±12\sqrt{10}}{2} when ± is plus. Add 60 to 12\sqrt{10}.

$$x=\frac{12\sqrt{10}+60}{2}$$

Divide 60+12\sqrt{10} by 2.

$$x=6\sqrt{10}+30$$

Now solve the equation x=\frac{60±12\sqrt{10}}{2} when ± is minus. Subtract 12\sqrt{10} from 60.

$$x=\frac{60-12\sqrt{10}}{2}$$

Divide 60-12\sqrt{10} by 2.

$$x=30-6\sqrt{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 30+6\sqrt{10} for x_{1} and 30-6\sqrt{10} for x_{2}.

$$x^{2}-60x+540=\left(x-\left(6\sqrt{10}+30\right)\right)\left(x-\left(30-6\sqrt{10}\right)\right)$$

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.

$$x ^ 2 -60x +540 = 0$$

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

$$r + s = 60 rs = 540$$

Two numbers r and s sum up to 60 exactly when the average of the two numbers is \frac{1}{2}*60 = 30. 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>

$$r = 30 - u s = 30 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 540

$$(30 - u) (30 + u) = 540$$

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

$$900 - u^2 = 540$$

Simplify the expression by subtracting 900 on both sides

$$-u^2 = 540-900 = -360$$

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

$$u^2 = 360 u = \pm\sqrt{360} = \pm \sqrt{360}$$

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

$$r =30 - \sqrt{360} = 11.026 s = 30 + \sqrt{360} = 48.974$$

Evaluate

$$x^{2}-60x+540$$