How do you solve #x^2 + 10x + 24 = 0# using completing the square?

3 Answers
Jun 14, 2015

Answer:

#0 = x^2+10x+24 = (x+5)^2 - 1#

Hence #x = -5+-sqrt(1) = -5+-1#

Explanation:

#(x+5)^2 = x^2+10x+25#

So we have:

#0 = x^2+10x+24 = (x+5)^2 - 1#

Add #1# to both ends to get:

#(x+5)^2 = 1#

So:

#x+5 = +-sqrt(1) = +-1#

Subtract #5# from both sides to get:

#x = -5+-1#

That is #x=-6# or #x=-4#

Jun 14, 2015

Answer:

Factor #y = x^2 + 10x + 24# by completing the square

Explanation:

#y = x^2 10x + (25 - 25) + 24 = 0#
#y = (x + 5)^2 - 1 = 0#
#(x + 5)^2 = 1# -> #x + 5 = +- 1#

x = -5 + 1 = -4
x = -5 - 1 = -6

Jun 14, 2015

Answer:

Create a perfect square trinomial on the left side of the equation, then factor it and solve for #x#. The general equation for a perfect square trinomial is #a^2+2ab+b^2=(a+b)^2#.

Explanation:

#x^2+10x+24=0#

We are going to create a perfect square trinomial on the left side of the equation, then solve for #x#. The general equation for a perfect square trinomial is #a^2+2ab+b^2=(a+b)^2#.

Subtract #24# from both sides.

#x^2+10x=-24#

Divide the coefficient of the #x# term by #2#, then square the result, and add it to both sides.

#10/2=5# #; 5^2=25#

#x^2+10x+25=-24+25# =

#x^2+10x+25=1#

We now have a perfect square trinomial on the left side, in which #a=x and b=5#. Factor the trinomial, then solve for #x#.

#(x+5)^2=1#

Take the square root of both sides.

#x+5=+-sqrt1# =

#x=+-sqrt1-5# =

#x=sqrt 1-5=1-5=-4#

#x=-sqrt1-5=-1-5= -6#

#x=-4#
#x=-6#