Solve t^2-10t-11

$$t ^ { 2 } - 10 t - 11$$

Factor

$$\left(t-11\right)\left(t+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-11. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-11\right)=-11$$

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. The only such pair is the system solution.

$$a=-11 b=1$$

Rewrite t^{2}-10t-11 as \left(t^{2}-11t\right)+\left(t-11\right).

$$\left(t^{2}-11t\right)+\left(t-11\right)$$

Factor out t in t^{2}-11t.

$$t\left(t-11\right)+t-11$$

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

$$\left(t-11\right)\left(t+1\right)$$

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.

$$t^{2}-10t-11=0$$

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.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-11\right)}}{2}$$

Square -10.

$$t=\frac{-\left(-10\right)±\sqrt{100-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$t=\frac{-\left(-10\right)±\sqrt{100+44}}{2}$$

Add 100 to 44.

$$t=\frac{-\left(-10\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$t=\frac{-\left(-10\right)±12}{2}$$

The opposite of -10 is 10.

$$t=\frac{10±12}{2}$$

Now solve the equation t=\frac{10±12}{2} when ± is plus. Add 10 to 12.

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

Divide 22 by 2.

$$t=11$$

Now solve the equation t=\frac{10±12}{2} when ± is minus. Subtract 12 from 10.

$$t=-\frac{2}{2}$$

Divide -2 by 2.

$$t=-1$$

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 -1 for x_{2}.

$$t^{2}-10t-11=\left(t-11\right)\left(t-\left(-1\right)\right)$$

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

$$t^{2}-10t-11=\left(t-11\right)\left(t+1\right)$$

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.

$$x ^ 2 -10x -11 = 0$$

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

$$r + s = 10 rs = -11$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -11

$$(5 - u) (5 + u) = -11$$

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

$$25 - u^2 = -11$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -11-25 = -36$$

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

$$u^2 = 36 u = \pm\sqrt{36} = \pm 6$$

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

$$r =5 - 6 = -1 s = 5 + 6 = 11$$

Evaluate

$$\left(t-11\right)\left(t+1\right)$$