2x^2+8x-3=02x2+8x−3=0
rArr color(blue)2(x^2+4x)=3⇒2(x2+4x)=3
If color(orange)(4x)4x is the middle term of a squared binomial with the form
color(white)("XXX")(x+a)^2 = (x^2+color(orange)(2ax)+a^2)XXX(x+a)2=(x2+2ax+a2)
then
color(white)("XXX")color(orange)a=color(orange)2XXXa=2
and
color(white)("XXX")color(green)(a^2)=color(green)(4)XXXa2=4 must be added to x^2+4xx2+4x to complete the square.
This will result in adding color(blue)2 xxcolor(green)42×4 to the left side of the equation;
so we must add this to the right side as well to keep the equation valid.
color(white)("XXX")color(blue)2(x^2+4x+color(green)4)=3+color(blue)2xxcolor(green)4XXX2(x2+4x+4)=3+2×4
Rewriting the parenthesized factor as a squared binomial and simplifying.
color(white)("XXX")2(x+2)^2=11XXX2(x+2)2=11
Now divide both sides by 22, leaving only a squared binomial on the left:
color(white)("XXX")(x+2)^2=11/2XXX(x+2)2=112
Take the square root of both sides (don't forget that there will be both a positive and negative root on the right side):
color(white)("XXX")x+2=+-sqrt(11/2)XXXx+2=±√112
Subtract 22 from both sides to isolate the variable xx
color(white)("XXX")x=-2+-sqrt(11/2)XXXx=−2±√112
