How do you solve #2x ^ { 2} - 20x - 6= 0#?

1 Answer
George C.
Dec 20, 2016

#x = 5+-2sqrt(7)#

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=2sqrt(7)# as follows:

#0 = 2x^2-20x-6#

#color(white)(0) = 2(x^2-10x-3)#

#color(white)(0) = 2(x^2-10x+25-28)#

#color(white)(0) = 2((x-5)^2-(2sqrt(7))^2)#

#color(white)(0) = 2((x-5)-2sqrt(7))((x-5)+2sqrt(7))#

#color(white)(0) = 2(x-5-2sqrt(7))(x-5+2sqrt(7))#

Hence:

#x = 5+-2sqrt(7)#

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