How do you solve using the completing the square method #x^2+24x+90=0#?

2 Answers
Apr 29, 2016

Answer:

#(x+12)^2 -54=0#

Explanation:

The expression #(x+a)^2# expands as #x^2 + 2ax + a^2#
so to complete the square we use half the coefficient of the middle term to be #a#.

We then subtract the equivalent of #a^2# (in this case #12^2#) and add the final term (#90#).

#(x+12)^2 -144+90#
#=(x+12)^2 -54#

Apr 29, 2016

Answer:

#=>x=-12+-3sqrt(6)" "# as exact values

#=> x~~-4.65" and "-19.35" "# to 2 decimal places

Explanation:

Standard for #" "y=ax^2+bx+c"#

Write as#" "y=a(x+b/(2a))^2 + c + (-b^2/(4a))#

The purpose of the #b^2/(4a)# is to mathematically remove an error that have been introduced by building #a(x+b/(2a))^2#

If you square #b/(2a)# then multiply it out by the variable #'a'# in front of the bracket you have introduced a value that was not in the original equation. So you remove it by subtraction.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "x^2+24x+90=0" "# Note that #a=1#

Write as:#" "y=(x+12)^2+90+k#

But #k=-(12)^2/4 = -144#

#color(brown)(" "=>y=(x+12)^2+90+k)color(blue)(" "->" "y=(x+12)^2-54)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x^2+24x+90=y=0=(x+12)^2-54#

So #(x+12)^2=+54#

#=>sqrt((x+12)^2)=sqrt(54)#

#=>x+12=+-sqrt(6xx3^2)#

#=>x=-12+-3sqrt(6)#

#=> x~~-4.65" and "-19.35# to 2 decimal places
Tony B