How do you solve #x^ { 2} - 10x + 16= 0 #?

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Answer 1 · 2017-10-16T02:27:50

#x=8,x=2#

The goal here is to find two numbers who product is #16# and sum is #-10#.

This requires a little of trial and error but eventually these two numbers are #-8# and #-2# since:

#-8times-2=16#

#-8+-2=-10#

So we now have

#(x-8)(x-2)=0#

Now we solve for #x# separately by writing the above equation into two separate equations:

#x-8=0->x=8#

#x-2=0->x=2#

Answer 2 · 2017-10-16T02:30:17

See a solution process below:

First, factor the left side of the equation as:

#(x - 2)(x - 8) = 0#

Now, solve each term on the left for #0# to find the solutions:

Solution 1:

#x - 2 = 0#

#x - 2 + color(red)(2) = 0 + color(red)(2)#

#x - 0 = 2#

#x = 2#

Solution 2:

#x - 8 = 0#

#x - 8 + color(red)(8) = 0 + color(red)(8)#

#x - 0 = 8#

#x = 8#

The Solutions Are: #x = 2# and #x = 8#