Solve x^2-150x+594

$$x ^ { 2 } - 150 x + 594$$

Factor

$$\left(x-\left(75-3\sqrt{559}\right)\right)\left(x-\left(3\sqrt{559}+75\right)\right)$$

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.

$$x^{2}-150x+594=0$$

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.

$$x=\frac{-\left(-150\right)±\sqrt{\left(-150\right)^{2}-4\times 594}}{2}$$

Square -150.

$$x=\frac{-\left(-150\right)±\sqrt{22500-4\times 594}}{2}$$

Multiply -4 times 594.

$$x=\frac{-\left(-150\right)±\sqrt{22500-2376}}{2}$$

Add 22500 to -2376.

$$x=\frac{-\left(-150\right)±\sqrt{20124}}{2}$$

Take the square root of 20124.

$$x=\frac{-\left(-150\right)±6\sqrt{559}}{2}$$

The opposite of -150 is 150.

$$x=\frac{150±6\sqrt{559}}{2}$$

Now solve the equation x=\frac{150±6\sqrt{559}}{2} when ± is plus. Add 150 to 6\sqrt{559}.

$$x=\frac{6\sqrt{559}+150}{2}$$

Divide 150+6\sqrt{559} by 2.

$$x=3\sqrt{559}+75$$

Now solve the equation x=\frac{150±6\sqrt{559}}{2} when ± is minus. Subtract 6\sqrt{559} from 150.

$$x=\frac{150-6\sqrt{559}}{2}$$

Divide 150-6\sqrt{559} by 2.

$$x=75-3\sqrt{559}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 75+3\sqrt{559} for x_{1} and 75-3\sqrt{559} for x_{2}.

$$x^{2}-150x+594=\left(x-\left(3\sqrt{559}+75\right)\right)\left(x-\left(75-3\sqrt{559}\right)\right)$$

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.

$$x ^ 2 -150x +594 = 0$$

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

$$r + s = 150 rs = 594$$

Two numbers r and s sum up to 150 exactly when the average of the two numbers is \frac{1}{2}*150 = 75. 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>

$$r = 75 - u s = 75 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 594

$$(75 - u) (75 + u) = 594$$

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

$$5625 - u^2 = 594$$

Simplify the expression by subtracting 5625 on both sides

$$-u^2 = 594-5625 = -5031$$

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

$$u^2 = 5031 u = \pm\sqrt{5031} = \pm \sqrt{5031}$$

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

$$r =75 - \sqrt{5031} = 4.070 s = 75 + \sqrt{5031} = 145.930$$

Evaluate

$$x^{2}-150x+594$$