Question #c3149

2 Answers
Jul 26, 2016

Answer:

#-3# and #-1#
or
#13# and #15#

Explanation:

two consecutive odd integers
#2x+1# and #2x+3#

the sum is
#2x+1+2x+3 =y#

product is
#27+6(y)=(2x+1)*(2x+3)#

or

#27+6(2x+1+2x+3)=(2x+1)*(2x+3)#

now we can solve for #x# and plug back into #2x+1# and #2x+3#
for the final answer

#27+24(x+1)=4x^2+8x+3#

#48+24x=4x^2+8x#

#48=4x^2-16x#

#0=x^2-4x-12#

#0=(x+2)(x-6)#

so #x=6# or #x=-2#so lets plug in and check

a) #-3# and #-1#

b) #13# and #15#

both check so we are done

Jul 26, 2016

Answer:

#13,15#

or

#-3,-1#

Explanation:

Let's break this problem down:

We have two consecutive odd integers. How can we express that an two integers are odd and consecutive?

Imagine we had some integer #n#. We don't know if this integer is even or odd, however. But, we can guarantee that we have an even integer if we say #2n#. Even if #n# is odd, #2n# is even. Thus, #2n+1# is guaranteed to be odd.

If #2n+1# is an odd integer, then we know there will be consecutive odd integers #2# away in either direction, that is: #(2n+1)-2=2n-1#, and #(2n+1)+2=2n+3#.

So, we can say that our two consecutive odd integers here are #bb(2n-1)# and #bb(2n+1)#.

Now, we need to set up an equation that represents the statement "their [the odd integers'] product is #27# more than #6# times their sum."

Let's start with the integers product. Product means multiplication, so the product of our odd integers is just:

#(2n-1)(2n+1)#

However, this product is #27# times greater than #6# times their sum. First write #6# times their sum, mathematically:

#6[(2n-1)+(2n+1)]#

And since the product is #27# times greater, we can set up the equation as follows:

#(2n-1)(2n+1)=27+6[(2n-1)+(2n+1)]#

Now we can solve for #n#.

First, add #(2n-1)+(2n+1)#:

#(2n-1)(2n+1)=27+6(4n)#

#(2n-1)(2n+1)=27+24n#

Distribute (FOIL) on the left-hand side. Notice that since it is in the form #(a+b)(a-b)=a^2-b^2#, we get:

#4n^2-1=24n+27#

Move all the terms to the left-hand side:

#4n^2-24n-28=0#

Divide each term by #4#.

#n^2-6n-7=0#

To factor this, we're looking for two integers whose sum is #-6# and product is #-7#. The integers that fit these criteria are #-7 # and #1#, so we see the factorization of:

#(n-7)(n+1)=0#

Implying that:

#n=7# or #n=-1#

We'll take the positive solution of #n=7#. Our two consecutive odd integers are #2n-1# and #2n+1#, so when #n=7#, that translates into #bb13# and #bb15#.

If we wish to accept negative answers with #n=-1#, we see that #2n-1=-3# and #2n+1=-1#.