Solve x^2-2x-99

$$x ^ { 2 } - 2 x - 99$$

Factor

$$\left(x-11\right)\left(x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-99. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-99\right)=-99$$

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 -99.

$$1,-99 3,-33 9,-11$$

Calculate the sum for each pair.

$$1-99=-98 3-33=-30 9-11=-2$$

The solution is the pair that gives sum -2.

$$a=-11 b=9$$

Rewrite x^{2}-2x-99 as \left(x^{2}-11x\right)+\left(9x-99\right).

$$\left(x^{2}-11x\right)+\left(9x-99\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-11\right)+9\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x+9\right)$$

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.

$$x^{2}-2x-99=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-99\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-99\right)}}{2}$$

Multiply -4 times -99.

$$x=\frac{-\left(-2\right)±\sqrt{4+396}}{2}$$

Add 4 to 396.

$$x=\frac{-\left(-2\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{-\left(-2\right)±20}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±20}{2}$$

Now solve the equation x=\frac{2±20}{2} when ± is plus. Add 2 to 20.

$$x=\frac{22}{2}$$

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{2±20}{2} when ± is minus. Subtract 20 from 2.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and -9 for x_{2}.

$$x^{2}-2x-99=\left(x-11\right)\left(x-\left(-9\right)\right)$$

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

$$x^{2}-2x-99=\left(x-11\right)\left(x+9\right)$$

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.

$$x ^ 2 -2x -99 = 0$$

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

$$r + s = 2 rs = -99$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -99

$$(1 - u) (1 + u) = -99$$

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

$$1 - u^2 = -99$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -99-1 = -100$$

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

$$u^2 = 100 u = \pm\sqrt{100} = \pm 10$$

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

$$r =1 - 10 = -9 s = 1 + 10 = 11$$

Evaluate

$$\left(x-11\right)\left(x+9\right)$$