What is the square root of ax^2+bx+c?

#sqrt(ax^2+bx+c)=?# i tried myself and got #a^(2)x+sqrt(c),a^(2)x+b/2# But i do not think That is exactly right.

1 Answer
Geoff K.
Feb 2, 2018

#sqrt(ax^2+bx+c)=sqrt a" "x +sqrt c#, as long as #a# and #c# aren't negative, and #b=+-2sqrt(ac).#

Explanation:

If #ax^2+bx+c# is a perfect square, then its square root is #px+q# for some #p# and #q# (in terms of #a, b, c#).

#ax^2+bx+c = (px+q)^2#
#color(white)(ax^2+bx+c) =p^2" "x^2 + 2pq" "x + q^2#

So, if we are given #a#, #b#, and #c#, we need #p# and #q# so that

#p^2=a#,
#2pq = b#, and
#q^2=c#.

Thus,

#p=+-sqrt a#,
#q=+-sqrt c#, and
#2pq = b#.

But wait, since #p= +-sqrta# and #q=+-sqrtc#, it must be that #2pq# is equal to #+-2sqrt(ac)# as well, so #ax^2+bx+c# will only be a perfect square when #b=+-2sqrt(ac).# (Also, in order to have a square root, #a# and #c# must both be #ge 0#.)

So,

#sqrt(ax^2+bx+c)=px+q#
#color(white)(sqrt(ax^2+bx+c))=sqrt a" "x +sqrt c#,

if

#a>=0#,
#c>=0#, and
#b=+-2sqrt(ac)#.

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