How do you solve #s^2-3(s+2)=4# ?

2 Answers
Shantelle
May 9, 2018

#s = 2 and s = -5#

Explanation:

First, use the distributive property to simplify #color(blue)(-3(s+2)#:

#(-3 * s) - (3 * 2)#

#-3s - 6#

So now the equation is:
#s^2 - 3s - 6 = 4#

Subtract #color(blue)4# from both sides to get one side to equal to #0#:
#s^2 - 3s - 6 quadcolor(blue)(-quad4)= 4 quadcolor(blue)(-quad4)#

#s^2 - 3s - 10 = 0#

This equation is now in standard form, or #ax^2 + bx + c = 0#.

To factor and solve for #s#, we need two numbers that:
#1.# Multiply up to #ac = 1(-10) = -10#
#2.# Add up to #b = -3#

The two numbers that do that are #color(blue)2# and #color(blue)(-5)#:
#1. quadquad 2 * -5 = -10#
#2. quadquad 2 - 5 = -3#

Therefore, we put it in factored form, or:
#(s-2)(s+5) = 0#

Since they multiply up to #0#, we can do:
#s-2 = 0 and s+5 = 0#
#quadquadquad# #s = 2 and quadquadquadquad s = -5#

Hope this helps!

Camilleon
May 9, 2018

Warning: Long answer, but hopefully worth it

s = -2 or 5

Explanation:

Following PEMDAS:

#s^2 - 3(s + 2) = 4#

First, let's distribute -3 to s and +2. Remember that distributing means you're multiplying -3 by both terms in the parentheses. You should now have:

#s^2 -3s - 6 = 4#

Now, because you have no like terms, add six to both sides. You should now have:

#s^2 - 3s = 10#

This a quadratics equation, and you need to set the equation to 0 in order to solve it. So, subtract 10 from both sides. You should now have:

#s^2 - 3s - 10 = 0#

Now, use the XBOX method. First, we need to multiply our first term by our last term #(s^2 * -10)#. We need then get #-10s^2#.

Now, you have to multiply 2 numbers that get you #-10s^2# but also add to #-3s#. To do this, factor out 10:

1 - 10
2 - 5

-5 and 2 multiply to get you -10, and add to -3, so these are the terms we want to use. You should now have:

#s^2 -5s + 2s - 10 = 0

Now, make a table like this:

                           ?                  ?
                     ?  #s^2#          -5s

                   ?    2s                  -10

See where the question marks are? You want to find out what multiplies to give you the terms, starting with #s^2#:

#s * s = s^2#, so these two will be s:

                            s                    ?
                     s  #s^2#          -5s

                   ?    2s                  -10

Now, you have two question marks remaining. Since you have s and ? that multiplies to -5s, the ? will be -5 because s * -5 = -5s. Add that in:

                            s                    -5
                     s  #s^2#          -5s

                   ?    2s                  -10

Now, we have one variable left. s * ? = 2s and -5 * ? equals -10. ? will be 2 because s * 2 = 2s and -5 * 2 = -10. So, insert your final variable:

                           s                    -5
                     s  #s^2#          -5s

                   2  2s                  -10

Now, your equation looks like this: (s + 2)(s - 5) = 0
Isolate each ordered pair and set it to 0 to find out what is s.

(s + 2) = 0; s = -2
(s - 5) = 0; s = 5

Source and for more info:

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