How do you find the roots for #x^2 – 14x – 32 = 0#?

2 Answers
Jun 6, 2015

In an equation of the following form

#ax^2+bx+c=0#

the method to find the roots is:

1) calculate #Delta = b^2-4ac#
2) if #Delta=0# there is only one root #x_0=(-b)/(2a)#
3) if #Delta>0# there are two roots #x_(-)= (-b-sqrt(Delta))/(2a)#
and #x_(+) = (-b+sqrt(Delta))/(2a)#
4) if #Delta<0# there is no real solution

Example:

#x^2-14x-32=0#

#rarr a=1; b=-14; c=-32#

#rarr Delta = (-14)^2 - 4 * 1 * (-32) = 196 +128 = 324#

#Delta>0# therefore we have two roots:

#x_(-) = (14-sqrt324)/2 = (14-18)/2 = -4/2 = -2#

#x_(+) = (14+sqrt324)/2 = (14+18)/2 = 32/2 = 16#

Let us check the validity of our results:

#(-2)^2-14*(-2)-32 = 4+28-32 = 0 rarr OK#

#(16)^2-14*(16)-32 = 256-224-32 = 0 rarr OK#

Jun 6, 2015

There are several methods we could use. Here's one.

Notice that #2*16=32# and the difference between 2 and 16 is 14.

So, if the signs work out, we can factor.

#x^2-14x-32=(x+2)(x-16)#

So, #x^2-14x-32=0# if and only if

#(x+2)(x-16)=0#

Thus, we need

#x+2=0# or #x-16=0#

The solutions are:

#x=-2#, #x=16#.