Does #x ^ { 2} - 4x - 2= 0# have any real solutions?

1 Answer
George C.
Jan 27, 2018

Yes, namely #x = 2+-sqrt(6)#

Explanation:

Given:

#x^2-4x-2 = 0#

Note that this is in standard form:

#ax^2+bx+c = 0#

with #a=1#, #b=-4# and #c=-2#

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-4))^2-4(color(blue)(1))(color(blue)(-2)) = 16+8 = 24 = 2^2*6#

Note that this is positive, but not a perfect square.

Hence we can deduce that the quadratic equation has irrational real roots.

One way of solving it is to use the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (4+-2sqrt(6))/2#

#color(white)(x) = 2+-sqrt(6)#

Alternatively we could complete the square and use the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A=(x-2)# and #B=sqrt(6)#, as follows:

#0 = x^2-4x-2#

#color(white)(0) = x^2-4x+4-6#

#color(white)(0) = (x-2)^2-(sqrt(6))^2#

#color(white)(0) = ((x-2)-sqrt(6))((x-2)+sqrt(6))#

#color(white)(0) = (x-2-sqrt(6))(x-2+sqrt(6))#

Hence:

#x = 2+-sqrt(6)#

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