Complete: #x^2+kx+25# is a perfect square then #k=.....# ?????

2 Answers
Jim G.
May 4, 2018

#k=+-10#

Explanation:

#"note that "(a+-b)^2larrcolor(blue)"perfect square"#

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

#"compare to "x^2+kx+25#

#"here "a=x" and "b=5#

#rArr2ab=k=2xx x xx5=10#

#"that is "(x+5)^2=x^2+10x+25#

Tony B
May 4, 2018

#k=+-10#

Explanation:

For this to be a perfect square we have:
#(sqrt(x^2)+sqrt(25))^2=x^2+kx+25#

As
#[color(white)(".")(-sqrt(25))xx(-sqrt(25))color(white)(".")] =[color(white)(".")(+sqrt(25))xx(+sqrt(25))color(white)(".")]=+25#

#color(blue)("What we should consider is:")#

#(sqrt(x^2)+-sqrt(25))^2=x^2+kx+25#

#color(white)("ddd")(x+-5)^2color(white)("ddd")=x^2+kx+25#

#color(blue)("Consider the case:")#

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

#x^2+10x+25=x^2+kx+25 color(white)("ddd")=>color(white)("ddd") k=+10#

#color(blue)("Consider the case:")#

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

#x^2-10x+25=x^2+kx+25 color(white)("ddd")=>color(white)("ddd") k=-10#

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