Solve x^2-x-90

$$x ^ { 2 } - x - 90$$

Factor

$$\left(x-10\right)\left(x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-90. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-90\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -1.

$$a=-10 b=9$$

Rewrite x^{2}-x-90 as \left(x^{2}-10x\right)+\left(9x-90\right).

$$\left(x^{2}-10x\right)+\left(9x-90\right)$$

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

$$x\left(x-10\right)+9\left(x-10\right)$$

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

$$\left(x-10\right)\left(x+9\right)$$

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.

$$x^{2}-x-90=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-90\right)}}{2}$$

Multiply -4 times -90.

$$x=\frac{-\left(-1\right)±\sqrt{1+360}}{2}$$

Add 1 to 360.

$$x=\frac{-\left(-1\right)±\sqrt{361}}{2}$$

Take the square root of 361.

$$x=\frac{-\left(-1\right)±19}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±19}{2}$$

Now solve the equation x=\frac{1±19}{2} when ± is plus. Add 1 to 19.

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

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{1±19}{2} when ± is minus. Subtract 19 from 1.

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

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 10 for x_{1} and -9 for x_{2}.

$$x^{2}-x-90=\left(x-10\right)\left(x-\left(-9\right)\right)$$

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

$$x^{2}-x-90=\left(x-10\right)\left(x+9\right)$$

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.

$$x ^ 2 -1x -90 = 0$$

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

$$r + s = 1 rs = -90$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -90

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -90$$

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

$$\frac{1}{4} - u^2 = -90$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -90-\frac{1}{4} = -\frac{361}{4}$$

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

$$u^2 = \frac{361}{4} u = \pm\sqrt{\frac{361}{4}} = \pm \frac{19}{2}$$

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

$$r =\frac{1}{2} - \frac{19}{2} = -9 s = \frac{1}{2} + \frac{19}{2} = 10$$

Evaluate

$$\left(x-10\right)\left(x+9\right)$$