Solve x^2-5x-50

$$x ^ { 2 } - 5 x - 50$$

Factor

$$\left(x-10\right)\left(x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-50. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-50\right)=-50$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -50.

$$1,-50 2,-25 5,-10$$

Calculate the sum for each pair.

$$1-50=-49 2-25=-23 5-10=-5$$

The solution is the pair that gives sum -5.

$$a=-10 b=5$$

Rewrite x^{2}-5x-50 as \left(x^{2}-10x\right)+\left(5x-50\right).

$$\left(x^{2}-10x\right)+\left(5x-50\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-10\right)+5\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+5\right)$$

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.

$$x^{2}-5x-50=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-50\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-50\right)}}{2}$$

Multiply -4 times -50.

$$x=\frac{-\left(-5\right)±\sqrt{25+200}}{2}$$

Add 25 to 200.

$$x=\frac{-\left(-5\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-\left(-5\right)±15}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±15}{2}$$

Now solve the equation x=\frac{5±15}{2} when ± is plus. Add 5 to 15.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{5±15}{2} when ± is minus. Subtract 15 from 5.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -5 for x_{2}.

$$x^{2}-5x-50=\left(x-10\right)\left(x-\left(-5\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}-5x-50=\left(x-10\right)\left(x+5\right)$$

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.

$$x ^ 2 -5x -50 = 0$$

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

$$r + s = 5 rs = -50$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{2}. 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>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -50

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -50$$

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

$$\frac{25}{4} - u^2 = -50$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -50-\frac{25}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =\frac{5}{2} - \frac{15}{2} = -5 s = \frac{5}{2} + \frac{15}{2} = 10$$

Evaluate

$$\left(x-10\right)\left(x+5\right)$$