How do you solve #x^2-10x-9=0#?

1 Answer
Dec 13, 2016

Answer:

#x = 5+-sqrt(34)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a = (x-5)# and #b=sqrt(34)# as follows:

#0 = x^2-10x-9#

#color(white)(0) = x^2-10x+25-34#

#color(white)(0) = (x-5)^2-(sqrt(34))^2#

#color(white)(0) = ((x-5)-sqrt(34))((x-5)+sqrt(34))#

#color(white)(0) = (x-5-sqrt(34))(x-5+sqrt(34))#

Hence #x = 5+-sqrt(34)#