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