Solve x^2-50x+600=

$$x ^ { 2 } - 50 x + 600 =$$

Factor

$$\left(x-30\right)\left(x-20\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+600. To find a and b, set up a system to be solved.

$$a+b=-50 ab=1\times 600=600$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 600.

$$-1,-600 -2,-300 -3,-200 -4,-150 -5,-120 -6,-100 -8,-75 -10,-60 -12,-50 -15,-40 -20,-30 -24,-25$$

Calculate the sum for each pair.

$$-1-600=-601 -2-300=-302 -3-200=-203 -4-150=-154 -5-120=-125 -6-100=-106 -8-75=-83 -10-60=-70 -12-50=-62 -15-40=-55 -20-30=-50 -24-25=-49$$

The solution is the pair that gives sum -50.

$$a=-30 b=-20$$

Rewrite x^{2}-50x+600 as \left(x^{2}-30x\right)+\left(-20x+600\right).

$$\left(x^{2}-30x\right)+\left(-20x+600\right)$$

Factor out x in the first and -20 in the second group.

$$x\left(x-30\right)-20\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x-20\right)$$

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.

$$x^{2}-50x+600=0$$

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.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 600}}{2}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 600}}{2}$$

Multiply -4 times 600.

$$x=\frac{-\left(-50\right)±\sqrt{2500-2400}}{2}$$

Add 2500 to -2400.

$$x=\frac{-\left(-50\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-\left(-50\right)±10}{2}$$

The opposite of -50 is 50.

$$x=\frac{50±10}{2}$$

Now solve the equation x=\frac{50±10}{2} when ± is plus. Add 50 to 10.

$$x=\frac{60}{2}$$

Divide 60 by 2.

$$x=30$$

Now solve the equation x=\frac{50±10}{2} when ± is minus. Subtract 10 from 50.

$$x=\frac{40}{2}$$

Divide 40 by 2.

$$x=20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 30 for x_{1} and 20 for x_{2}.

$$x^{2}-50x+600=\left(x-30\right)\left(x-20\right)$$

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.

$$x ^ 2 -50x +600 = 0$$

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

$$r + s = 50 rs = 600$$

Two numbers r and s sum up to 50 exactly when the average of the two numbers is \frac{1}{2}*50 = 25. 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>

$$r = 25 - u s = 25 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 600

$$(25 - u) (25 + u) = 600$$

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

$$625 - u^2 = 600$$

Simplify the expression by subtracting 625 on both sides

$$-u^2 = 600-625 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =25 - 5 = 20 s = 25 + 5 = 30$$

Evaluate

$$\left(x-30\right)\left(x-20\right)$$