How do you solve #0 = x^2 + 10x + 5#?

2 Answers
George C.
May 12, 2016

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

Explanation:

Complete the square and use the difference of squares identity:

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

with #a=(x+5)# and #b=2sqrt(5)# as follows:

#0 = x^2+10x+5#

#=(x+5)^2-25+5#

#=(x+5)^2-20#

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

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

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

Hence:

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

Alan P.
May 12, 2016

#x=-5+2sqrt(5)color(white)("XXX")orcolor(white)("XXX")x=-5-2sqrt(5)#

Explanation:

The easiest way for this particular example is to use the quadratic formula:

For an equation of the form #color(red)(a)x^2+color(blue)(b)x+color(green)(c)=0#
the solutions are given by #x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)(a))#

For the given example

#color(red)(a)=1#
#color(blue)b=10# and
#color(green)c=5#

So
#color(white)("XXX")x=(-10+-sqrt(10^2-4(1)(5)))/(2(1))#

#color(white)("XXXX")=(-10+-4sqrt(5))/2#

#color(white)("XXXX")=-5+-2sqrt(5)#

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