How do you solve #x^2-11x+19=-5@?

1 Answer
Mar 27, 2015

A quadratic equation is simply another way of solving a problem if the solution cannot be factored logically.

First we can start with some quick review:

Let’s say we have the equation #x^(2)+ 2x - 3# for example. This equation could be solved logically using the factors of the first and last terms.

To begin, we can state the factors of the first term, #x^(2)#. Imagine there’s an invisible 1 in front of the #x^(2)#, therefore the factors are 1, because only #1 * 1, or -1*-1# will multiply to get one. Then we can analyze the third term, #-3#. The factors of #-3# are either #1 * -3, or -1 * 3#.

Now we can check and see if any of the factors can combine in order to get a #+2#, the middle term (don’t worry about the x’s, those will carry over). Recall #1= -1, 1#, and #-3 = 1, -1, 3, -3#

From our factors we can use a -1 and a 3 to get +2. Therefore,
#(x+3)(x-1)=0# is our derived factorization. Then plug in the values to make the statement true, -3 or 1 will both result in an answer of 0 and our the possible values for x.

However , when the logical factorization seen above is not possible, we can plug our numbers into the quadratic equation .

#ax^(2)+bx+c# is the standard way we view an equation. Using the values from the equation above, #a= 1, b=2, and c=-3#.

After our a, b, and c values are found we can plug them into the actual quadratic equation.

#(-b+-sqrt(b^(2)-4ac))/(2a)#

Note : This equation may look intimidating, but as long as you follow factoring rules, you should have no problem. It’s totally normal to come out with an answer containing square roots.